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IN  MEMORIAM 
FLORIAN  CAJORl 


riEST  COURSE  IN  ALGEBRA 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/firstcourseinalgOOwheerich 


FIEST    COURSE 


IN 


ALGEBRA 

BY 

ALBERT   HARRY   WHEELER 

Teacher  op  Mathematics  in  the  English  High  School  at  Worcester, 
Massachusetts 


WITH    EIGHT    THOUSAND    EXAMPLES 

INCLUDING 

THREE  THOUSAND  MENTAL  EXERCISES 

\ 


BOSTON 
LITTLE,  BROWN,  AND   COMPANY 

1907 


Copyri'jht,  1007, 
By  Little,  Brown,  and  Company. 


THE  UNIVERSITY   PRESS,    CAMBKIPGK,   U.  S.  A. 


\Aj'+-'^ 


PREFACE 

rilHIS  book  may  be  used  by  students  who  have  completed  a 
■^  course  in  arithmetic  but  who  have  not  previously  studied 
algebra.  The  statements  of  all  principles  and  the  explana- 
tions of  all  examples  have  been  made  as  simple  and  direct  as 
possible.  Emphasis  has  been  placed  upon  the  reason  for  each 
step  taken,  whether  it  be  in  the  proof  of  a  principle  or  in  the 
explanation  of  an  exercise.  Such  proofs  of  principles  as  may 
be  omitted  by  the  student  when  taking  the  subject  for  the  first 
time  have  been  plainly  marked. 

The  examples  are  all  new  and  have  not  been  copied  from 
other  text-books.  There  are  4,465  exercises  for  written  solu- 
tion, 3,301  for  mental  solution,  and  423  explanatory  examples, 
—  a  total  of  8,189  examples. 

These  examples  have  been  so  constructed  and  graded  as  to 
contain  a  great  variety  of  number-combinations.  Thus  the 
student  is  constantly  drilled  in  arithmetic. 

The  writer  first  used  mental  exercises  in  the  class-room 
in  1895,  and  has  constantly  employed  them  since  that  time 
with  the  exception  of  the  years  1897-99,  which  were  spent  in 
graduate  study  at  Clark  University.  The  first  ten  or  fifteen 
minutes  of  each  recitation  period  are  commonly  devoted  by 
his  pupils  to  the  mental  solution  of  a  large  number  of  simple 
drill  problems.  In  this  way  all  of  the  pupils  have  an  oppor- 
tunity to  recite  several  times  during  every  recitation  period. 


VI  PREFACE 

After  they  gain  confidence  and  a  certain  amount  of  skill  in 
solving  the  mental  exercises,  it  is  found  that  the  more  difficult 
problems  which  are  given  for  written  solution  are  undertaken 
with  readiness. 

By  the  use  of  the  mental  exercises  the  teacher  is  able  to 
gain  a  better  knowledge  of  the  progress  of  the  pupil  than  by 
giving  many  written  examinations,  and  there  is  the  advantage 
that  mistakes  on  the  part  of  the  pupil  can  be  corrected  imme- 
diately. Written  examinations  show  the  teacher  the  way  the 
pupil  has  thought,  but  mental  exercises  show  the  pupil  the 
way  to  think  in  the  future. 

The  applied  problems  are  concerned  with  subjects  of  modern 
interest.  In  particular,  problems  have  been  introduced  illus- 
trating the  applications  of  certain  familiar  laws  of  physics, 
such  as  those  relating  to  the  lever,  falling  bodies,  expansion 
of  gases,  etc.  The  traditional  problems  which  involve  un- 
natural and  absurd  situations  have  been  excluded. 

The  graphs  were  drawn  by  the  writer,  and  will  be  found  to 
be  accurate.  The  explanations  of  the  examples  have  been 
given  in  such  a  way  that  all  reference  to  graphs  may  be  omit- 
ted if  the  time  devoted  to  the  subject  is  found  to  be  insufficient 
for  their  consideration. 

A  system  of  numerical  checks  is  used  throughout  the  book, 
and  the  pupil  is  constantly  encouraged  to  test  results  obtained 
rather  than  to  depend  upon  the  authority  of  the  teacher  or  of 
a  printed  list  of  answers. 

In  the  development  of  the  subject  the  distinction  between 
natural  forms  of  number  and  "  artificial "  or  invented  forms 
of  number  has  been  constantly  kept  in  view,  and  by  means  of 
the  Principle  of  No  Exception  the  necessity  has  been  shown 


PREFACE  vii 

for  the  invention  of    negative  number    and    other   forms  of 
"artificial"  number. 

Simple  proofs  and  illustrations  of  the  principles  of  equiva- 
lence of  equations  have  been  given,  and  the  distinction  be- 
tween identical  and  conditional  equalities  has  been  carefully 
pointed  out. 

Attention  has  been  given  to  detail  in  the  classification  and 
arrangement  of  the  subject-matter  for  the  purpose  of  making 
it  easily  available  for  reference  and  of  simplifying  the  presen- 
tation of  the  subject. 

ALBERT   HARRY   WHEELER. 

Worcester,  Mass. 

April,  1907. 


Publishers*  Note.  —  The  Brief  Edition  of  this  book,  which   takes 
the  pupil  as  far  as  Quadratics,  contains  6,327  examples. 


TABLE  OF  CONTENTS 


CHAPTER  I 

First  Ideas 

Pac« 

Letters  may  be  usefl  to  represent  numbers 1 

Symbols  of  number  and  of  operation  carried  over  from  arithmetic  to  algebra  1 

Definitions  of  certain  words  which  constitute  part  of  the  language  of  als^ebra  4 

Algebraic  functions 5 

Distinction  between  Conditional  Equations  and  Identical  Equations     ...  7 

Numerical  Substitutions 7 

Axioms 9 

The  Principle  of  Substitution 9 


NUMERICAL    EXPRESSIONS 

CHAPTER  II 

An  Extension  of  the  Idea  of  Number 

The  Principle  of  No  Exception 12 

I'ositive  and  Nejjfative  Quantities 13 

Invention  of  Negative  Numbers 18 

CHAPTER  III 

Fundamental  Laws  of  Algebra 

Laws  of  Commutation  and  Association  for  the  Addition  and  Subtraction  of 

Positive  and  Negative  Numbers 23 

Steps  on  a  Line 24 

Law  of  Signs 28 

CHAPTER  IV 

Addition  and  Subtraction  of   Positive  and  Negative  Numbers 

Symbols  of  Grouping 29 

I.    Addition 32 

II.    Subtraction 37 

Correspondence  between  the  Addition  and  Subtraction  of  Positive  and  Nega- 
tive Numbers 39 


X  TABLE  OF   CONTENTS 

CHAPTER  V 
Multiplication  and  Division  of  Positive  and  Negative  Numbers 

I.   MULTIPLICATION  Page 

Definitions 45 

An  extended  definition  of  multiplication 46 

Law  of  Signs  for  Multiplication 49 

Coninuitative  Law  for  Multiplication 50 

Associative  Law  for  Multiplication 52 

Distributive  Law  for  Multiplication 52 

Zero  as  a  Factor 53 

Principles  Relating  to  Powers 55 

II.    DIVISION 

Definitions       ....    * 57 

Zero  cannot  be  used  as  a  divisor 58 

Law  of  Signs  for  Division • 59 

An  extended  definition  of  division 60 

Commutative  Law  for  Division 61 

Associative  Law  for  Division 62 

Principles  Governing  the  Removal  and  Insertion  of  Parentheses 62 

Distributive  Law  for  Division 65 

lilTERAL    EXPRESSIONS 

OPERATIONS  WITH    INTEGRAL   ALGEBRAIC    EXPRESSIONS 

CHAPTER  VI 

Addition  and  Subtraction 

Definitions       69 

Addition  of  Monomials 73 

(a)  Addition  of  Dissimilar  Terms 73 

(b)  Addition  of  Similar  Terms 74 

Subtraction  of  Monomials 77 

Reduction  of  a  Polynomial  to  Simplest  Form 80 

The  Check  of  Arbitrary  Values 81 

Addition  of  Polynomials 81 

Detached  Coefficients 82 

Subtraction  of  Polynomials       84 

CHAPTER  VII 

Multiplication 

Principles  Relating  to  Powers 88 

Product  of  Powers  of  the  Same  Base 88 

Powers  of  Products  of  Different  Bases 89 

Product  of  Two  or  More  Monomials 9' 


TABLE  OF   CONTENTS  xi 

Pagr 

Multiplication  of  a  Polynomial  by  a  Monomial 94 

Multiplication  of  One  Polynomial  by  Another 96 

Detached  Coefficients 98 

Homogeneity  as  a  Check  upon  Accuracy 99 

liemoval  of  Parentheses 101 

Standard  Identities     .     ." 102 

Square  of  a  Binomial  Sum 102 

Square  of  a  Binomial  Difference 103 

Multiplication  of  the  Sum  of  Two  Numbers  by  their  Difference    .     .     104 

Square  of  a  Polynomial 105 

Product  of  Two  Binomials  of  the  Forms  J"  +  a  and  a: +  6 107 

Product  of  Two  Binomials  of  the  Forms  aa:  + 6  and  ca:  +  rf    ....     108 
Powers  of  a  Binomial Ill 

CHAPTER   VIII 

Division 

Law  of  Exponents 112 

One  Power  Divided  by  another  Power  of  the  Same  Base    .......  113 

Division  of  One  Monomial  by  Another 114 

Division  of  a  Polynomial  by  a  Monomial 116 

Division  of  One  Polynomial  by  Another 118 

Development  of  the  Process 119 

The  Division  Transformation 124 

Detached  Coefficients  in  DivisioD 127 

Numerical  Checks  in  Division 127 

Remainder  Theorem 129 

Synthetic  Division 130 

Factor  Theorem 132 

Applications  of  the  Remainder  Theorem 135 

Law  of  Polynomial  Quotients 135 

CHAPTER  IX 

Graphical  Representation  of  the  Variation 
OF  Functions  of  a  Single  Variable 

Variation  of  a  Function 137 

Specification  of  Points  in  a  Plane 138 

Locating  Points  by  Coordinates 139 

Typical  Graphs 143 

Inverse  Use  of  Graphs 147 

CHAPTER  X 

General  Principles  Governing  the  Transformation 
OF  Algebraic  Equations 

Identical  Equations 149 

Conditional  Equations 152 

Equivalent  Equations 155 


xii  TABLE  OF  CONTENTS 

Page 

General  Principles  Governing  Transformations  of  Conditional  Equations     .  156 

I.  Identical  Substitution 156 

II.  Addition  and  Subtraction 156 

Applications : 

(i.)  Transposition  of  Terms 157 

(ii.)  Suppression  of  Identical  Terms 157 

(iii.)  Reversal  of  the  Sign  of  every  Term 158 

(iv.)  Reduction  to  the  Standard  Form -.1  =  0 158 

m.  Multiplication 161 

Principle  Relating  to  Extra  Roots 162 

IV.  Division .  163 

Principle  Relating  to  Loss  of  Roots *   .     .  163 

Applications : 

(i.)  Freeing  the  members  of  an  equation  of  fractional 

coefficients 164 

(ii.)  Transformation  so  that  a  particular  term  shall  have  a 

specified  coefficient 165 

The  Process  of  Derivation  is  not  always  Reversible 165 

CHAPTER  XI 
Equations  of  the  First  Degree  Containing  One  Unknown 

Solution  of  a  Linear  Equation 168 

Suggestions  concerning  the  solution  of  linear  equations 171 

Algebraic  Expression 175 

Simple  Problems  Involving  One  Unknown  Quantity 177 

Problems  in  Science 183 

CHAPTER  XII 
Factors  of  Rational  Integral  Expressions 

The  Problem  of  Factoring 185 

Definitions 185 

Expressions  in  which  all  of  the  terms  contain  a  common  monomial  factor   .  186 

Expressions  in  which  groups  of  terms  have  a  common  factor 188 

Type      I.  Trinomial  Squares 190 

Type    IL  The  Difference  of  Two  Squares 196 

Expressions  of  the  form  x*  ±  hx^y^  -\-  y^ 200 

Type  III.  Trinomials  of  the  type  x^  +  sx  -{- p 201 

Trinomials  of  the  form  a:2"* +  sx*-f/3 204 

Trinomials  of  the  form  x^  -f  sxy  +  py^ 205 

Tvpe  IV.  Trinomials  of  the  type  ax'^  -\- hx  +  c 206 

First  Method 206 

Second  or  Trial  Method 207 

Type   V.  Binomial  Sums  and  Differences 212 

Polvnomials  of  at  least  the  third  degree  with  reference  to  some  letter,  x     .  214 

Suggestions  Relating  to  Methods 216 

Application  of  the  Principles  of  Factoring  to  the  Solution  of  Equations      .  223 

Principle  of  Equivalence 224 


TABLE   OF   CONTENTS  Xlll 

CHAPTER  XIII 
Highest  Common  Factor 

Page 

Definitions 228 

Highest  Common  Factor  by  Factoring 229 

Monomial  Expressions       229 

Polynomial  Expressions 231 

Highest  Common  Factor  of  Polynomials 233 

Principles  Governing  the  Process 233 

Development  of  the  Process 235 

Highest  Common  Factor  of  Three  or  More  Expressions 243 

CHAPTER  XIV 
Lowest  Common  Multiple 

Definitions 245 

Lowest  Common  Multiple  by  Inspection 255 

Lowest  Common  Multiple  by  means  of  Highest  Common  Factor     ....  247 

Lowest  Common  Multiple  of  Three  or  More  Expressions 248 

Mental  Review  Exercise 249 

CHAPTER  XV 

Rational  Fhactions 

Extension  of  the  Idea  of  Number  by  the  Principle  of  No  Exception    .     .     .  251 

Definitions ' 252 

Principles  Relating  to  the  Signs  of  a  Fraction 254 

Reduction  to  Lowest  Terms 257 

Reduction  of  Improper  Fractions  to  Integral  or  Mixed  Expressions    .     .     .  261 
Reduction  of  Fractions  to  Equivalent  Fractions  having  a  Common  Denom- 
inator     262 

Addition  and  Subtraction  of  Fractions 265 

Reduction  of  Integral  or  Mixed  Expressions  to  Fractional  Forms  ....  269 
Principles  Fundamental  to  the  Processes  of  Multiplication  and  Division  In- 
volving Fractions 270 

Multiplication  of  an  Integer  by  a  Fractional  Multiplier 272 

Multiplication  of  One  Fraction  by  Another 272 

Power  of  a  Fraction 275 

Division  of  a  Whole  Number  by  a  Fraction 277 

Division  of  One  Fraction  by  Another 278 

Direct  Process 278 

Indirect  Process.     (For  the  operation  of  division  by  a  fraction  may 
be  substituted   that  of   multiplication  by  the  reciprocal   of  the 

divisor) 279 

Division  of  One  Fraction  by  Another 280 

Complex  Fractions , 283 

Continued  Fractions 286 


xiv  TABLE  OF  CONTENTS 

Paob 

Indeterminate  Forms 290 

Interpretation  of  the  Indeterminate  Forms  q'  q'  oo ' °         •     •  292 

Factors  of  Fractional  Expressions 295 

Mental  Review  Exercise 296 

CHAPTER  XVI 

Fractional  and  Literal  Equations 

fractional  equations 

Principle  of  Equivalence 298 

General  Directions  for  Solving  Fractional  Equations 302 

LITERAL   EQUATIONS 

Integral  Literal  Equations 306 

Literal  Equations  as  Formulas 308 

Numerical  Checks  for  the  Solutions  of  Literal  Equations 310 

Fractional  Literal  Equations 312 

Problems  Solved  by  Fractional  Equations 316 

General  Problems 318 

The  Interpretation  of  Solutions  of  Problems 321 

Problems  in  Physics 323 

Review  Exercise 326 

CHAPTER  XVII 

Simultaneous  Linear  Equations 

general  principles  of  equivalence 

Definitions 327 

Graph  of  a  Linear  Equation  containing  Two  Unknowns         328 

Classification  of  Pairs  of  Kquations 329 

Independent  Equations 330 

Consistent  Equations 330 

Inconsistent  Equations 331 

Equivalent  Equations 332 

Equivalent  Systems  of  Equations 333 

solutions   of    SYSTEMS   OF    SIMULTANEOUS    EQUATIONS 

Elimination 334 

General  Principles 334 

I.  Elimination  by  Substitution 335 

Systems  of  Linear  Equations  containing  Two  Unknowns  ....  836 

Elimination  by  Comparison 339 

II.  Elimination  by  Addition  or  Subtraction 341 

General  Solution  of  a  System  of  Two  Consistent,  Independent,  Linear  Equa- 
tions containing  Two  Unknown  Numbers 34G 

Systems  of  Linear  Equations  containing  Three  or  More  Unknowns     ,     .     .  347 

Graphical  Record  of  the  Process  of  Solution           348 

Systems  of  Fractional  K(iuations  Solved  Like  Equations  of  the  First  Degree  355 

Problems  Involving  Simultaneous  Equations 359 


TABLE   OF   CONTENTS  XY 

CHAPTER  XVIII 

Evolution 

Page 

Definitions 366 

Rational  or  Cominensurablfe  Numbers  and  Irrational  or  Incommensurable 

Numbers 367 

General  Principles  Governing  Root  Extraction 368 

The  Principal  Root  of  a  Number 369 

Roots  of  Monomials 371 

Principles  Governing  Operations  with  Radical  Symbols 371 

Root  of  a  Power 371 

Root  of  a  Product 371 

Root  of  a  Root 372 

Power  of  a  Root 373 

Root  of  a  Quotient 374 

Square  Roots  of  Polynomials 375 

Cul»e  Roots  of  Polynomials 378 

Square  Roots  of  Arithmetic  Numbers 381 

Cube  Roots  of  Arithmetic  Numbers 387 

CHAPTER  XIX 

Theory  of  Exponents 

Extension  of  the  Meaning  of  Exponent 391 

Fundamental  Index  Laws 391 

Interpretation  of  Zero  and  Unity  as  Exponents 392 

A  Negative  Integer  as  an  Exponent 393 

A  Positive  Fraction  as  an  Exponent 396 

A  Negative  Fraction  as  an  Exponent 398 

Products  of  Powers  and  Quotients  of  Powers 402 

Powers  of  Powers 404 

1  'owers  of  Products  and  Powers  of  Quotients 406 

CHAPTER  XX 
Irrational  Numbers  and  the  Arithmetic  Theory  of  Surds 

I.    IRRATIONAL    NUMBERS 

Commensurable  and  Incommensurable  Numbers 411 

Representation  of  Irrational  Numbers  by  their  Approximate  Values  .     .    .  412 

II.    ARITHMETIC   THEORY   OF   SURDS 

Definitions 415 

Reduction  of  Surds  to  Simplest  Form , 416 

I.  The  Radicand  an  Integer 417 

II.  The  Radicand  a  Fraction .....419 

HI.  Reduction  of  a  Surd  to  an  Equivalent  Surd  of  Lower  Order     .     .  422 


XVI  TABLE  OF  CONTENTS 

Pagb 

Addition  and  Subtraction  of  Surds 423 

Reduction  to  Equivalent  Forms 425 

Change  of  Order 427 

Multiplication  of  Surds 428 

Multiplication  of  Polynomials  Involving  Surds 430 

Involution  of  Surds 432 

Division  of  Surds 432 

Rationalization 434 

Factors  Involving  Surds 439 

Evolution  of  Surds 441 

Properties  of  Quadratic  Surds 442 

Square  Root  of  a  Binomial  Quadratic  Surd       444 

CHAPTER  XXI 
Imaginary  and  Complex  Numbers 

I.    IMAGINARY    NUMBERS 
The  "  Principle  of  No  Exception,"  applied,  leads  to  the  Invention  of  Im- 
aginary Numbers 446 

Multiplication  by  t       447 

Powers  of  I 447 

Division  by  j 448 

Addition  and  Subtraction  of  Imaginary  Numbers 450 

Multiplication  of  Imaginary  Numbers 451 

Division  of  Imaginary  Numbers 453 

The  Operation  of  "  Realizing"  the  Denominator  of  a  Fraction 453 

Graphical  Representation  of  Imaginary  Nujnbers 456 

IT.    COMPLEX   NUMBERS 

Addition  and  Subtraction  of  Complex  Numbers 459 

Multiplication  of  Complex  Numbers 459 

Division  of  Complex  Numbers 459 

Complex  Factors  of  Rational  Integral  Expressions 462 

Square  Root  of  Complex  Numbers 462 

Graphical  Representation  of  Complex  Numbers 464 

The  Number  System  of  Algebra  Complete 466 

Mental  Review  Exercise 467 

CHAPTER   XXII 

Equations  of  the  Second  Degree  Containing  One  Unknown 

Definitions 469 

Standard  Quadratic  Equation  ax^  -\-  hx  -\-  c  =  0 469 

Graphs  of  Quadratic  Equations 470 

I.  Incomplete  Quadratic  Equations 472 

First  Method.     Solution  by  Factoring 473 

Second  Method.     Solution  by  Extracting  the  Square  Roots  ....  474 


TABLE   OF   CONTENTS  xvu 

Page 

II.  Complete  Quadratic  Equations 477 

Solution  by  Factoring 477 

Solution  by  Completing  the  Square 478 

Completion  of  the  Square  with  respect  to  a  given  Binomial      .  479 

Hindu  Method 485 

General  Solution 487 

Rational  Fractional  Equations  containing  One  Unknown 491 

Principle  Relating  to  Extra  Roots 491 

Solution  of  Formulas  for  Specified  Letters 502 

Solution  of  Problems 503 

Problems  in  Physics 509 

CHAPTER  XXIII 
Thkory  of  Quadratic  Equations  Containing  One  Unknown 

Nature  of  the  Roots 514 

The  Discriminant 514 

Relations  Between  the  Roots  and  Coefficients 517 

[    Formation  of  an  Equation  having  Specified  Roots 519 

One  Root  Known 522 

Mental  Review  Exercise 523 

CHAPTER  XXIV 

Irrational  Equations  and  Special  Equations 
Containing  a  Single  Unknown 

IRRATIONAL    EQUATIONS 

Principle  Relating  to  Extra  Roots 527 

P 

.    Equations  of  the  form  x^  =  a 535 

^  SPECIAL    EQUATIONS 

Equations  which  are  Quadratic  with  Reference  to  some  Particular  Power 

of  the  Unknown 538 

liquations  which  are  Quadratic  with  Reference  to  some  Expression  Contain- 
ing the  Unknown 541 

Binomial  Equations 545 

Reciprocal  Equations 545 

Problems  in  Physics 547 

Review  Exercise 548 

CHAPTER  XXV 

Systems  of  Simultaneous  Equations 
Involving  Quadratic  Equations 

systems  of  two  equations  containing  two  unknowns 

I     The  Eliminant 551 

Number  of  Solutions 552 


xviii  TABLE  OF  CONTENTS 

Page 

I.  Elimination  by  Substitution 553 

Graphical  Interpretation 554 

II.  Reduction  of  Systems  of  E(iuations  by  Factoring 558 

Graphical  Interpretation 560 

III.  Systems  of  Two  Honiogeneous  Equations  of  the  Second  Degree  with 

Reference  to  Two  Unknowns 562 

Solution  by  Factoring 562 

Solution  by  Expressing  the  Value  of  One  Unknown  as  a  Multiple  of 

the  Other 566 

IV.  Reduction  of  Systems  of  Equations  by  Division 570 

V.  Systems  of  Symmetric  Equations 572 

Solutions  by  Special  Devices 577 

System    of    Three  or   More  Equations  Containing   Three   or   More 

Unknowns 584 

Problems 590 


CHAPTER  XXVI 
Ratio,  Proportion,  and  Variation 

I.  RATIO 

Definitions 594 

Approximate  Value  of  an  Incommensurable  Ratio 596 

II.  PROPORTION 

General  Principles 599 

Problems  in  Physics 605 

III.  variation 

One  Independent  Variable 609 

Fundamental  Principle 610 

Two  or  More  Independent  Variables 611 

General  Principles 612 

Problems  in  Physics 614 

Mental  Review  Exercise 615 


CHAPTER  XXVII 

The  Progressions 

The  Sequence 619 

I.  arithmetic  progression 

Arithmetic  Progression 620 

Arithmetic  Means 622 

Series 623 

Sum  of  the  Terms  of  an  Arithmetic  Progression 624 

Problems  in  Physics 627 


TABLE   OF   CONTENTS  XIX 

II.    HARMONIC    PROGRESSION  Pagk 

Harmonic  Progression 627 

Harmonic  Means 628 

III.  GEOMETRIC    PROGRESSION 

Geometric  Progression 630 

Geometric  Means 632 

Sum  of  the  'J'erms  of  a  Geometric  Progression 634 

Sum  of  tlie  Terms  of  an  Infinite  Decreasing  Geometric  Progression  .     .     .  636 

Generating  Fraction  of  a  Repeating  Decimal  Fraction 638 

Evaluation  of  a  Repeating  Decimal  Fraction 639 

Review  Exercise 641 

CHAPTER   XXVIII 

The  Binomial  Theorem 

The  Binomial  Theorem  for  Positive  Integral  Exponents 643 

Multiplication   and   Division   Rule   for   Calculating  the  Coefficients  Suc- 
cessively     643 

Proof  of  the  Binomial  Theorem  for  Positive  Integral  Exponents  by  Mathe- 
matical Induction 647 

The  Binomial  Theorem  for  Negative  or  Fractional  Exponents 649 

Selection  of  a  Particular  Term  in  the  Expansion  of  the  Power  of  a  Binomial  651 

The  Factorial  Notation 651 

Binomial  Coefficients 656 

Aj)plication  of  the  Binomial  Theorem  to  the  Extraction  of  Roots  of  Arith- 
metic Numbers 656 

Mental  Review  Exercise 657 

Review  Exercise 658 


INDEX 661 


FIRST  COURSE  IN  ALGEBRA 


CHAPTER  I 

FIRST  IDEAS 

1.  The  student  of  mathematics  who  undertakes  the  study  of 
algebra  will  find  at  the  outset  that  the  new  science  is  but  arith- 
metic in  a  different  form  and  under  another  name. 

2.  In  algebra  numbers  are  represented  by  letters,  and  while  defi- 
nite values  are  not  assigned  to  the  different  letters  employed,  they 
are  used  with  the  idea  that  each  letter  represents  some  numerical 
value,  and  that  different  letters  commonly  represent  different 
values. 

3.  Letters  which  stand  for  numbers  whose  values  remain  fixed 
and  unchanged,  either  throughout  the  discussion  of  a  particular 
problem  or  for  all  time,  are  called  constants,  and  those  whose 
values  may  be  supposed  to  change  during  the  discussion  of  a  prob- 
lem are  called  variables. 

Letters  which  represent  numbers  whose  values  are  known  are 
called  knowns,  and  those  whose  values  may  for  the  moment  be 
unknown  are  commonly  called  unknowns. 

It  is  customary  in  elementary  algebra  to  represent  constants  by 
the  first  few  letters  of  the  alphabet,  «,  ft,  c,  and  the  unknown  values, 
and  also  variables,  by  the  last  few  letters,  ^,  ?/,  z. 

In  many  applications  of  algebra  to  formulas  of  physics  and  other 
sciences,  or  even  in  certain  algebraic  expressions,  any  letter  or  group 
of  letters  may  be  chosen  as  representing  variables  or  unknowns ;  in 
which  case  the  remaining  letters,  if  there  be  any,  may  be  regarded 
as  constants,  or  knowns. 

4.  The  symbols  used  to  denote  the  operations  of  addition, 
subtraction,  multiplication,  division,  and  root  extraction  in  arith- 
metic are  used  in  the  same  sense  in  algebra. 


2  FIRST  COURSE   IN  ALGEBRA 

5.  The  symbol  for  addition,  +,  is  read  "  plus,"  and  for  subtraction, 
— ,  is  read  "minus."  These  symbols  indicate  that  the  numbers  be- 
fore which  they  are  placed  are  to  be  added  to  or  subtracted  from  the 
numbers  which  immediately  precede  them. 

Thus,  the  order  of  operations  is  from  left  to  right. 

For  example,  a  -\-  b,  read  "a  plus  ^>,"  means  that  some  number 
represented  by  ^  is  to  be  added  to  some  number  represented  by  a. 
Again,  x  —  y^  read  ".r  minus  v^,"  means  that,  if  x  and  ?/  represent 
numbers,  the  number  represented  by  y  is  to  be  subtracted  from  that 
represented  by  x, 

6.  Multiplication  may  be  indicated  in  several  ways;  by  the 
symbol  X  ;  by  a  dot  written  higher  than  a  decimal  point  placed 
between  the  multiplicand  and  multiplier;  or,  when  no  possible  am- 
biguity can  arise,  by  simply  writing  the  multiplicand  and  multiplier 
consecutively. 

Thus,  if  a  and  h  are  taken  to  represent  any  two  numbers,  their 
product  may  be  written  as  a  X  b,  a  •  b,  or  simply  as  a  b. 

In  each  case  the  product  is  read  either  "a  multiplied  by  6,"  or 
"rtr  times  ^." 

The  sign  for  multiplication  is  usually  omitted  between  two  letters, 
or  between  a  number  and  a  letter. 

Thus,  a  b  means  a  X  b;  Amn  means  4:  X  m  X  n. 

The  sign  for  multiplication  cannot  be  omitted  between  numerals 
in  arithmetic  because  of  the  positional  system  of  notation  employed. 

Thus,  72  must  be  regarded  as  standing  for  seventy- two  wherever 
it  is  found,  and  not  as  7  X  2,  or  14. 

To  avoid  confusion  with  the  decimal  point,  the  dot,  when  used  as 
a  symbol  for  multiplication,  should  be  written  somewhat  higher. 

Thus,  7  •  2  may  be  taken  as  indicating  the  product  of  7  and  2, 
while  7.2  will  be  interpreted  as  a  decimal,  7  plus  2  tenths. 

7.  The  s3mabol  for  division  is  -f- .  Thus,  a  —  b,  read  "  a  divided 
by  b"  means  that  some  number  represented  by  a  is  to  be  divided 
by  another  number  represented  by  b.  The  number  a  is  regarded 
as  the  dividend  and  the  number  b  as  the  divisor,  as  in  arithmetic. 

As  alternative  notations  for  division  we  have  the  fractional 

notation,  j ,  the  ratio  notation,  a  :  b,  and  the  soliclus  notar 
tion,  a/b. 


FIRST   IDEAS  8 

8.  In  an  unbroken  chain  of  additions  and  subtractions,  or  in  an 
unbroken  chain  of  multiplications  and  divisions,  the  operations  are 
to  be  carried  out  successively  from  left  to  right. 

Thus,  from  the  chain  of  additions  3  +  4  +  5  +  6  we  obtain  18, 
by  performing  the  operations  successively. 

Again,  performing  the  operations  of  the  chain  25  —  8  —  6  —  3 
successively  from  left  to  right,  we  obtain  as  a  result  8. 

The  result  obtained  by  performing  successively  the  multiplications 
of  the  chain  2  X  -4  X  8  is  64. 

The  number  resulting  from  performing  the  divisions  of  the  chain 
8-7-4-7-2,  successively  from  left  to  right,  is  1. 

9.  However,  in  a  chain  of  operations  containing  additions,  sub- 
tractions, multiplications  and  divisions  together,  the  mult'qdicatmis 
and  dkisions  must  he  performed  first  in  the  order  as  indicated,  and 
then  the  operations  of  addition  and  subtraction  in  their  order. 

Thus,  to  find  the  number  represented  by  the  chain  of  operations 
5  +  12  X  2  -r-  3  —  6  -f-  3  +  1,  we  may  proceed  as  follows  : 

Multiplying  12  by  2,  and  dividing  the  product  24  by  3,  we  obtain 
8  ;  also,  dividing  6  by  3,  we  have  as  a  quotient  2. 

Hence  the  above  chain  of  operations  reduces  to  the  chain 
5  +  8  —  2+1. 

Performing  these  last  additions  and  subtractions  successively  from 
left  to  right,  we  obtain  12  as  a  result. 

Exercise  I.    1. 
Find  the  values  of  the  following : 


1. 

5  +  2X3. 

11. 

1  +  2X3  +  4X5. 

2. 

10  -  4  X  2. 

12. 

5  +  4X3-^2-l. 

3. 

12  +  10-^2. 

13. 

125  -^  25  —  25  -h  5  +  2. 

4. 

15  —  10-^5. 

14. 

5X5  —  4X4-3X3. 

5. 

7  +  6  +  5X4. 

15. 

7  X  8  -^  14  +  3  X  2  -^  3. 

6. 

7  +  6X5  +  4. 

16. 

8  -^  2  X  4  —  4 -^  2  X  6  +  12. 

7. 

7X6  +  5X4. 

17. 

6x3-^2-6X2-^3  +  7. 

8. 

8X5  —  6X3. 

18. 

6-^2X4-4^2X6  +  6X4-f-2. 

9. 

20-T-4  +  28^7. 

19. 

2X3X4  +  3X4X2+4X2X3. 

0. 

8-^2x3-^4+l. 

20. 

2X4X8-8-f-4-T-2  +  4-f-2X8. 

21.    20 -f- 4  X 

5  -  20  -^  5  X  4  -  20  -^  4  -r-  5. 

4  FIRST  COURSE  IN  ALGEBRA 

10.  Anything  which  can  be  multiplied  or  divided,  —  that  is, 
which  can  be  increased,  separated  into  parts,  or  measured,  is  called 
quantity. 

E.  g.  A  line  is  a  quantity  because  it  can  be  doubled,  tripled,  halved, 
etc.,  and  its  length  can  be  expressed  numerically  in  terms  of  another  line 
of  definite  length,  such  as  a  foot  or  a  yard,  taken  as  a  unit  of  measure. 

Weight  is  a  quantity,  since  it  can  be  measured  in  pounds,  ounces, 
grams,  etc. 

Time  is  a  species  of  quantity  whose  measure  can  be  expressed  in  terms 
of  seconds,  minutes,  hours,  etc. 

Color  is  not  a  quantity,  for  we  cannot  say  that  one  color  is  twice  as  great, 
or  one-half  as  great  as  another. 

The  operations  of  the  mind,  such  as  thought,  choice,  etc.,  are  not  quan- 
tities, for  they  are  incapable  of  measurement,  —  that  is,  of  direct  numerical 
comparison. 

11.  Any  letter  or  number  upon  which  an  operation  is  to  be  per- 
formed is  called  an  operand. 

12.  Any  combination  of  numbers,  letters,  and  symbols  of  opera- 
tion which  may  be  taken,  according  to  the  Principles  of  Algebra,  to 
represent  a  number,  is  called  an  algebraic  expression. 

13.  Parentheses  (  )  may  be  used  to  denote  that  an  algebraic 
expression  enclosed  by  them  is  to  be  treated  as  a  whole  throughout 
a  calculation. 

Thus,  3  X  (4  -f  5)  means  3  X  9,  or  27  ;  also  (4  +  1)  (7  -f  2) 
means  5  X  9,  or  45.     (See  Chapter  III,  §  3,  also  Chapter  IV,  §  1.) 

14.  When  two  numbers  are  multiplied  together  the  result  is 
called  the  product,  and  each  number  is  called  a  factor  of  the 
product. 

Thus,  12  is  the  product  of  4  and  3. 

15.  By  a  continued  product  is  meant  a  product  composed  of 
three  or  more  factors. 

Thus,  the  continued  product  of  2,  3  and  4  is  24. 

16.  An  exponent  is  a  small  number  placed  at  the  right  of  and 
a  little  above  any  number  or  factor  to  indicate  the  number  of  times 
that  number  or  factor  appears  as  a  factor  of  a  given  product. 

Thus,  3^  read  "  3  to  the  second  power "  or  "  3  square,"  means 
3X3;  2*,  read  "  2  to  the  fourth  power,"  means  2X2X2X2; 


FIRST   IDEAS  5 

a%^,  read  ''a  to  the  second  power,  times  h  to  the  third  power,"  or 
"a  square,  times  b  cube,"  means  aXaXbxbxb. 

When  no  exponent  is  written,  the  exponent  1  is  understood. 

Thus,  a:  means  a:^. 

17.  We  shall,  at  the  beginning  of  the  subject,  use  the  terms  sum, 
difference,  remainder,  product,  quotient,  etc.,  with  the  same  meanings 
in  algebra  as  in  arithmetic. 

18.  A  whole  number,  or  an  integer,  is  one  or  a  sum  of 
ones. 

Thus,  the  whole  number,  or  integer,  5  is  the  sum  of  five  I's. 

19.  The  numerical  value  of  an  algebraic  expression  for  speci- 
fied values  of  the  letters  appearing  in  it,  is  defined  for  the  present 
as  the  number  obtained  by  substituting  for  the  letters  given  nu- 
merical values,  and  performing  the  indicated  operations. 

Thus,  if  X  and  y  represent  2  and  3  respectively,  the  numeri- 
cal value  of  ^?/  is  6  ;  of  o^  +  ?/  is  5  ;  oi  x^  +  if  is  13  ;  and  that  of 
5;r  —  2  ?/  is  4. 

It  is  to  be  understood  that  such  values  only  are  to  be  given  to 
the  letters  as  will  allow  the  operations  to  be  performed.  The  neces- 
sary restrictions  on  the  values  of  the  letters  will  be  explained  as  they 
appear  in  later  chapters. 

20.  An  expression  is  said  to  be  a  function  of  some  specified 
letter  appearing  in  it  if  a  change  in  the  value  of  the  letter  produces 
in  general  a  change  in  the  value  of  the  expression. 

An  expression  which  is  a  function  of  x  may  be  indicated  by  writ- 
mgfix),  read  "function  a:*." 

The  expression  "function  x  "  suggests  to  us,  depends  for  its  valiie 
ujxm  the  value  of  x. 

E.  g.  The  distance  passed  over  by  a  person  in  a  given  time  depends  upon 
the  rate  at  which  the  person  travels,  and  may  be  spoken  of  as  being  a  "  func- 
tion "  of  the  rate. 

The  time  required  to  build  a  house  depends,  among  other  things,  upon 
the  number  of  workmen  employed,  and  may  be  said  to  be  a  "  function  "  of 
the  number  of  workmen. 

The  length  of  a  bar  of  metal  depends  upon  its  temperature,  and  we 
may  regard  the  length  of  a  particular  bar  as  being  a  "function"  of  the 
ti-mperature. 


6  FIRST  COURSE  IN  ALGEBRA 

21.  Although  different  Dumbers  in  arithmetic  are  represented  by 
different  definite  number  symbols,  each  number  may  be  represented 
in  an  unlimited  number  of  ways  by  combinations  of  other  numbers. 

Thus,  the  number  24  may  be  represented  by  12  X  2,  6X4, 
8  X  3,  20  +  4,  100  -r-  5  +  4,  etc. 

22.  Fixing  our  attention  upon  the  results  of  indicated  operations, 
as  in  the  illustration  above,  we  commonly  speak  of  such  expressions 
as  12  X  2,  6X4,  20  +  4,  etc.,  as  numbers,  meaning  thereby  the 
numbers  resulting  when  we  perform  these  operations. 

We  say  that  the  number  12  X  2  is  equal  to  the  number  20  +  4, 
since  each  represents  24  X  1. 

23.  As  a  symbol  of  relation  we  have  in  algebra,  as  in  arithmetic, 
the  sign  of  equality  =,  which  may  be  read  "equals,"  "is  equal  to," 
"  is  replaceable  by,"  etc. 

Thus,  8  +  2  =  10. 

24.  The  statement  in  symbols  that  two  expressions  represent  the 
same  number  is  called  an  equatiou. 

E.g.   5  +  1  =  4  +  2. 

The  part  at  the  left  of  the  equality  sign  is"  called  the  first  mem- 
ber, and  the  part  at  the  right  the  second  member,  of  the  equation. 

E.  g.  In  5  +  1  =  4  +  2,  5  +  1  is  the  first  member,  and  4  +  2  is  the  second 
member  of  the  equatiou. 

25.  TTie  sign  =  should  never  be  used  except  to  connect  numbers  or 
expressions  which  are  equal,  that  is,  which  stand  for  the  same  number. 
It  should  never  be  used  in  place  of  any  form  of  the  verb  "  to  6e." 

E.  g.    We  should  write,  "  Ans.  is  8,"  never  "  Ans.  =  8." 

26.  Two  algebraic  expressions  are  equivalent  when  they  repre- 
sent the  same  numerical  value,  no  matter  what  particular  values 
may  be  assigned  to  the  letters  appearing  in  them. 

E. g.   3 ic  +  ic  is  equivalent  to  4a;;  5^  —  2?/ is  equivalent  to  3 y. 

27.  An  equality  whose  members  are  equivalent  expressions  is 
called  an  identity. 

In  general  any  values  may  be  assigned  at  will  to  the  letters  ap- 
pearing in  an  identity.  Such  restrictions  as  may  be  necessary  in 
certain  cases  will  be  explained  in  a  later  chapter. 


FIRST   IDEAS  7 

28.  An  equality  which  is  true  for  particular  values  only  of  certain 
of  the  letters  appearing  in  it  is  called  a  conditional  equation. 

E.  g.  The  equation  a;  +  I  =  5  is  a  conditional  equation,  for  the  first 
member  is  equivalent  to  the  second  only  on  condition  that  x  be  given  the 
particular  value  4. 

We  shall  use  the  word  equation  to  mean  conditional  equation, 
that  is,  an  equality  which  is  not  an  identity. 

29.  In  order  to  distinguish  identical  from  conditional  equations 
we  shall  use  the  triple  sign  of  equality,  =  (read  "is  the  same  as," 
"is  identical  with,"  "stands  for,")  for  identical  equations,  and  the 
double  sign  of  equality  =  for  conditional  equations. 

To  conform  to  the  usage  in  arithmetic  we  shall  commonly  use  the 
double  sign  of  equality  =  histead  of  the  triple  sign  =  when  writing 
identities  in  which  arithmetic  numbers  only  appear. 

Other  reasons  for  this  use  of  the  sign  will  be  given  in  a  later 
chapter. 

Thus,  instead  of  6  +  2  =  8,  we  shall  write  6  +  2  =  8. 

30.  The  signs  >  and  <,  read  "is  greater  than,"  and  "is  less 
than,"  respectively,  are  used  as  symbols  of  inequality. 

E.  g.  We  may  denote  that  10  is  greater  than  8  by  writing  10  >  8 ;  that 
7  is  less  than  9  by  writing  7  <  9. 

It  should  be  noted  that  the  larger  end  of  each  symbol  is  directed 
toward  the  greater  quantity. 

31.  The  signs  of  equality  or  of  inequality,  when  crossed  by  lines, 
are  understood  as  meaning  "not  equal  to,"  "not  greater  than,"  and 
"not  less  than." 

E.  g.  2  ^f:  3  means  that  2  is  not  equal  to  3;  6  j/*  9  means  that  6  is  not 
greater  than  9 ;  8  ^  7  means  that  8  is  not  less  than  7. 

Exercise  I.    2 

Find  the  values  represented  by  the  following  expressions  when 
the  given  numerical  values  are  substituted  for  the  letters  : 
If  a  =  4,  ^  =  2,  c  =  5,  and  d  =  \^  find  the  value  of 

1.  a-{-  b.  4.   b  —  d.  7.    5  cd. 

2.  a  -{■  c.  5.    6  a.  8.   abc. 

3.  c-d,  6.    Sbc.  9.   a%. 


8  FIRvST  COURSE   IN   ALGEBRA 

10.  2al^.  12.   a^-^-ab.  14.    2cd-\-ahc. 

11.  3aW.  13.    (a  +  %.  15.    {a  +  h){c  ^- d). 

If  a  =  2,  6  =  5,  c  =  1,  6?  =  3,  find  the  numerical  values  repre- 
sented by  the  following  expressions: 

16.  ab  4-  he  +  cc?  +  da,  20.  (a  +  c)(6+</)  +  («  +  c?)(6+c). 

17.  ac  —  hd  +  ab  —  cd.  21.  (a  +  />)(^— c)+(6  +  c)(c— </). 

18.  abc  +  bed  -h  cda  +  a^.  22.  (a—b){c—d)  +  (b—c)(d—a). 

19.  (a4-6)(6+c)  +  (6+c)(c+rf).    23.  (a  +  6)c+(6+c)6?+(c+c^)a. 

If  rt  =  6,  ^  =  3,  .r  =  7,  and  y  =  1,  find  the  values  represented  by 
the  following  expressions : 

24.  a^  +  61  28.    (^^  +  />=)^'  +  (^-^  +  f)b. 

25.  a^  -  b\  29.   a*^  +  ^//>  -f  61 

26.  a'*  +  6  +  ^'2  ^  ^.  30.   a%  +  ah'^  -  x  —  y. 

27.  (rt  +  by  +  «  +  6.  31.   a^r'^  —  6/  —  a-  +  y. 

Verify  the  following  algebraic  identities  for  particular  values  of 
the  letters  appearing  in  them,  by  assigning  values  to  the  letters  : 
Ex.  32.     {a  +  \){a  +  2)  =  ^^  +  3a  +  2. 

If  the  members  are  identical  they  must  represent  equal  numbers  for  all 
values  which  may  be  assigned  to  a. 

Accordingly,  letting  a  =  4,  we  obtain,  substituting  4  for  a, 

(4  +  1)(4  +  2)  =  42 +  3x4  +  2 
5x6=  16 +  12 +  2 
30  =  30. 

Accordingly  the  identity  is  true  for  a  =  4.     By  substitution  it  will  be 
found  to  be  true  for  all  other  values  which  may  be  assigned  to  a. 

33.  (3^+5)(y+3)=y(^  +  8)  +  15.   35.  (2/+ 5)^  =  /+ 5  (2  7/  + 5). 

34.  {x-\-  ^y  =  x{x+  6)  +  9.         36.  w(?w  +  4)  +  4  =  ?wH4(»z+l). 

Ex.  37.    {a  +  by  =  a''  ^-2ab  +  h\ 

Assigning  to  a  and  h  the  values  8  and  2,  we  obtain  by  substitution, 

(8  +  2)2  =  82  +  2  X  8  X  2  +  22 
100  =  64  +  32  +  4 
100  B  100. 

Ex.  38.    {x  +  d){x  +  6)  =  ^'  +  (a  +  b)x  ■\-  ah. 

Substituting  4,  3,  and  2  for  a;,  a,  and  6  respectively,  we  obtain 


FIRST  IDEAS  9 

(4  +  3)(4  +  2)  =  42  +  (3  +  2)4  +  3  X  2 
7x6=  16 +  20 +  6 
42  =  42. 

S9.  (^  ■\-  yY  ^  4t:  xy  =  {x  —  yY. 

40.  {a'  +  ^>'0(^'  +  /)  -  {ax  +  hyY  =  (ay  -  hx)\ 

41.  {aP'  +  /)(«'  +  ^2)  =  {ax  -  hyY  +  {bx  +  aj/)^. 

42.  (.^  4-  y)*  —  ^*  —  /  =  3  ^j/(.r  +  ?/). 

43.  a'  +  (a^  +  «6  +  6^)^  =  (a^  +  ^2)  [^^  +  («  +  ^)']. 

44.  Or  +  3/)*  =  2  {x'  +  3^^)(^  +  ^)^  -  {x^  -  fY. 

45.  {a""  +  //)(c=^  +  ^•^)  =  {ac  +  /^af)^  +  (ad  -  bcY. 

46.  a*  +  ^'  =  («  +  h){a^  -  ab  +  ^^2). 

47.  («  -  ^)'  +  3  ab{a  -  b)  =  {a  +  bY  -  3  ab{a  +  b)  -  2  b\ 

48.  {x  +  yY  -  {'^  -  y)\x  +  2/)  =  4  xy{x  +  ?/). 

49.  {a  +  2)2  -  4  (a  +  1)2  +  6^2  -  4  («  -  1)^  +  {a  -  2)^  =  0. 

50.  {x-\- yY  =  x» -\-'dx'y -\-^xy''-[-y\ 

32.  An  axiom  is  the  statement  of  a  truth  which  may  be  inferred 
directly  from  our  experience,  or  from  the  nature  of  the  things 
considered. 

To  be  regarded  as  an  axiom,  a  truth  must  be  such  that  it  is  in- 
capable of  proof  further  than  its  mere  statement. 

As  axioms  common  to  mathematics,  we  may  state  the  following, 
which  were  called  by  an  early  writer  on  mathematics  Common 
Truths  about  Things : 

1.  Any  number  is  equal  to  itself. 
E.  g.   4  =  4. 

2.  The  Principle  of  Substitution.  The  numerical  value  of 
a  mathematical  expression  is  not  altered  when  for  any  number  or 
expression  in  it  we  substitute  an  equal  number  oi'  expression. 

That  is,  the  "form "  of  an  expression  may  be  changed  without 
altering  its  value. 

E.  g.  The  value  of  4  +  3  +  ^-^  remains  unaltered  if  for  ^  we  substitute 
its  equal  value  2;  or  again,  if  we  substitute  7  for  the  sum  of  4  and  3  and 
write  7  +  2. 

3.  Numbers  equal  to  the  same  number  are  equal. 
E.  g.    If  2  a;  =  a  and  10  =  a,  then  2  «;  =  10. 


10  FIRST  COURSE  IN  ALGEBRA 

4.  If  equal  numbers  be  added  to  equal  numbers  the  resulting  num- 
bers will  be  equal, 

E.  g.   If  a;  =  y,  then  a;  +  3  =  y  +  3. 

5.  If  equal  numbers  be  subtracted  from  equal  numbers  the  resulting 
numbers  ivillbe  equal. 

E.  g.    If  X  +  3  =  10,  then  x  =  10  -  3,  or  7. 

6.  If  equal  numbers  be  multiplied  by  equal  numbers  the  resulting 
numbers  will  be  equal. 

E.  g.    If  ^x  =  5,  then  two  times  \x  equals  two  times  5,  that  is,  x  =  10. 

7.  If  equal  numbers  be  divide  by  equal  numbers  {except  zero)  the 
resulting  numbers  will  be  equal. 

E.  g.  If  3x  =  12,  then  3 x  divided  by  3  equals  12  divided  by  3,  that  is, 
x  =  4. 

8.  Like  roots  of  equal  numbers  are  equal. 

E.  g.   If  x2  =  52,  then  x  =  5. 

There  are  two  square  roots,  three  cube  roots,  four  fourth  roots,  etc.,  of 
any  number,  so  that  when  applying  this  axiom  it  is  necessary  to  distinguish 
carefully  between  these  roots.  (See  Chapter  XVIII,  Principal  Values  of 
Roots.) 

33.  Substituting  the  word  "  identical "  for  the  word  "  equal  "  in 
each  of  the  statements  above,  we  have  corresponding  axiomatic  prin- 
ciples governing  identical  expressions. 

34.  If  .4  =  B  we  may  immediately  write  B  =  A,  since  this  is 
only  another  way  of  saying  the  same  thing. 

Identities  such  as  those  above,  which  are  formed  by  interchanging 
the  members,  are  said  to  be  one  the  converse  of  the  other. 

Ex.  1.    Find  the  value  which  must  be  assigned  to  a  in  order  that  the 
conditional  equation  2  a  +  3  =  15  may  be  true. 
Subtracting  3  from  each  member  we  obtain 

2a +  3 -3=  15  -3 
Hence  2  a  =  12. 

Dividing  both  members  of  the  last  equation  by  2,  we  obtain  finally  a  =  6. 
This  value  is  found  to  satisfy  the  original  conditional  equation. 


FIRST  IDEAS  11 

Exercise  I.    3 

Find,  by  appl3ring  the  axioms,  the  values  which  must  be  assigned 
to  the  letters  in  order  that  the  following  conditional  equations  may 
be  true. 

In  each  case  the  axiom  applied  should  be  stated,  and  the  result 
obtained  should  be  verified,  by  substituting  for  the  letter  in  the 
given  equation  the  value  found. 


1. 

6  +  4  =  10. 

11. 

l^d+  1  =  17 

2. 

c  -  2  =  7. 

12. 

hh-5  =  2. 

3. 

d—S  =  L 

13. 

I^J-1=6. 

4. 

2m  =  10. 

14. 

^z  =  8. 

5. 

6  w  =  42. 

15. 

fa  =  12. 

6. 

i^  =  8. 

16. 

i  /^  +  1  =  22. 

7. 

hl/=l'o. 

17. 

|c  +  3  =  13. 

8. 

4a+  1  =  13. 

18. 

1^-2  =  4. 

9. 

5  ^  +  2  =  22. 

19. 

fa  +  6  =  14. 

0. 

6c- 7  =  11. 

20. 

f  ^  -  1  =  5. 

12  FIRST  COURSE   IN  ALGEBRA 


CHAPTER  II 

AN  EXTENSION  OF  THE  IDEA  OF  NUMBER 

1.  In  arithmetic  we  have  found  it  possible  to  subtract  one  num- 
ber from  another  only  when  the  number  subtracted  was  not  greater 
than  the  number  from  which  it  was  taken. 

Such  combinations  of  numbers  as  6  —  9,  10  —  11,  etc.,  are  from  the  point 
of  view  of  arithmetic  wholly  destitute  of  meaning,  since  there  exists  no 
number,  that  is,  no  result  of  counting,  which  when  added  to  9  gives  6,  or 
when  added  to  11  gives  the  sum  10. 

Since  such  combinations  of  numbers  occur  frequently  in  mathematical 
work,  it  l)ecomes  necessary  for  us  to  give  them  a  meaning  if  we  are  to  allow 
them  to  remain  in  our  calculations.  To  do  this  we  find  it  necessary  to 
extend  our  notion  of  number.  The  combination  of  numbers  6  —  9,  as 
written,  suggests  to  us  a  diflFei-ence,  and  it  will  be  convenient  for  us  to  reckon 
with  it  as  with  every  other  "real"  or  "actual"  difference,  such  as  9  —  6,  or 
8  —  3,  etc.,  that  is,  a  diflFerence  in  which  the  subtrahend  is  less  than  the 
minuend. 

Principle  of  No  Exception 

2.  Mathematicians  are  accustomed  to  apply  the  names  of  familiar 
combinations  of  numbers  and  symbols  which  have  recognized  mean- 
ings to  all  similar  combinations,  even  when  these  do  not  appear  at 
first  to  admit  of  meaning,  or  even  to  make  sense.  This  principle, 
that  the  old  laivs  of  reckoning  and  the  old  meanings  must  he  carried 
over  to  include  all  .special  cases  of  a  given  general  type,  even  those 
which  may  appear  at  first  to  he  exceptions,  will  appear  under  many 
different  forms  throughout  the  whole  science  of  mathematics,  and 
wiU  be  referred  to  as  the  Principle  of  No  Exception. 

Instead  of  being  an  unwarranted  stretching  of  language,  as  it  may  appear 
at  first,  the  Principle  of  No  Exception  insists  rather  on  a  stretching  or 
broadening  of  ideas  to  fit  the  language  used,  in  order  to  avoid  contradictions 
which  might  otherwise  arise. 


AN   EXTENSION  OF   THE   IDEA  OF   NUMBER  13 

In  Arithmetic  the  primary  idea  of  a  fraction  is  a  "  part  of  unity  "  or  a 
"  broken  number." 

Thus,  f ,  ^,  \^,  are  fractions  in  this  sense. 

In  the  course  of  arithmetic  work,  combinations  appear  such  as  |^,  f,  ^, 
etc.,  which  look  like  fractions,  and  behave  like  fractions,  but  which  are  not 
in  the  original  sense  "  broken  numbers.  "  They  are  not  properly  fractions, 
and  are  accordingly  called  "  improper  fractions.  " 

The  Principle  of  No  Exception  is  then  applied,  and  such  combinations  as 
h  f'  t'  6'  ^^^'t  ^^^  ^^^»  without  exception,  spoken  of  as  fractions,  without 
specifying  whether  they  are  proper  or  improper  fractions,  so-called. 

3.  It  will  now  be  shown  that  the  application  of  this  idea  of  No 
Exception  leads  us  to  an  extension  of  our  previous  notions  concern- 
ing number,  and  to  the  invention  of  a  new  kind  of  number^  a  kind  of 
number  which  does  not  appear,  as  did  the  primary  numbers,  as  a 
result  of  counting,  but  which  nevertheless  may  be  used  in  our  cal- 
culations in  such  a  way  as  always  to  give  sense. 

« 

Positive  and  Negative  Quantities 

4.  Certain  words,  such  as 

forward  —  backward,  profit  —  loss, 

upward  —  downward,  earning  —  spending, 

north  —  south,  increasing  —  diminishing, 

rising  —  falling,  positive  —  negative, 

suggest  to  us  a  condition  of  two  things  such  that  each  tends  to 
destroy  the  effect  produced  by  the  other.  One  tends  to  increase 
whatever  the  other  tends  to  decrease.  The  terms  are  merely  relative 
and  imply  that,  from  some  point  of  view,  one  thing  tends  to  oppose 
another. 

E.  g.  If  travelling  east  takes  us  away  from  some  particular  place,  then 
from  the  same  point  of  view,  travelling  west  will  take  us  toward  that  same 
place. 

In  trade,  the  effect  produced  by  profits  offsets  the  effect  produced  by 


5.  Without  multiplying  illustrations  we  will  remark  simply  that 
the  terms  positive  and  negative  are  used  in  mathematics  in  such  a 
way  as  to  imply  that  there  is  some  opposition  such  that  if,  in  a 
calculation,  the  things  denoted  as  positive  should  be  added,  then 


14  FIRST  COURSE  IN  ALGEBRA 

those  called  negative  should  be  subtracted.  This  may  be  due  either 
to  the  nature  of  the  things  considered,  or  to  the  point  of  view  from 
which  we  regard  them. 

6.  The  opposition  between  two  sets  of  things  is  often  such  that  it 
is  of  no  consequence  which  is  considered  as  positive.  The  selection 
being  once  made,  so  long  as  the  things  of  one  set  in  a  calculation 
are  considered  as  positive  quantities,  those  of  the  other  set  must  in 
opposition  remain  as  negative  quantities. 

7.  By  the  absolute  value  of  a  quantity  expressed  in  terms  of 
some  unit  of  the  same  kind,  is  meant  the  number  of  times  the  unit 
is  contained  in  the  given  quantity.  This  is  without  regard  to  the 
quality  of  either  the  quantity  or  the  unit,  that  is,  as  to  whether 
both  are  positive  or  both  negative. 

8.  If  two  quantities  are  such  that,  when  combined  or  considered 
as  parts  of  one  whole,  any  given  amount  of  one  destroys  the  effect 
produced  by  an  amount  of  the  other  equal  in  absolute  value  to  that 
of  the  first,  these  two  quantities  are  called  opposites. 

In  mathematical  calculations  one  of  two  opposite  quantities  is 
called  positive  and  the  remaining  one  negative. 

9.  If,  in  any  calculation,  we  choose  to  regard  some  quantity  as 
being  positive, .  then  all  other  quantities  which  tend  to  increase  it 
must  be  considered  as  positive  also,  and  all  those  which  tend  to 
diminish  it  must  be  taken  as  negative. 

It  is  merely  a  matter  of  choice  which  one  of  two  opposite  quantities  is 
regarded  as  positive.  On  one  occasion  we  may  regard  motion  in  one  direc- 
tion, say  toward  the  right,  as  being  positive,  and  on  another  we  may  equally 
as  well  choose  to  regard  motion  toward  the  left  as  being  positive.  In  either 
case  motion  in  a  direction  directly  opposite  to  that  chosen  as  positive  would 
be  considered  as  negative  motion. 

Also,  if  we  choose  to  call  the  capital  invested  in  a  business  positive,  then 
all  profits  will  be  positive,  since  they  may  be  added  to  and  used  to  increase 
the  capital  ;  all  losses  and  expenses  will  be  negative,  for  they  tend  to 
diminish  the  capital,  since  they  must  be  subtracted  from  it. 

10.  It  is  not  essential  to  positive  quantities  that  they  be  numeri- 
cally greater  than  those  which  are  negative.  Thus,  losses  in  busi- 
ness, regarded  as  negative  quantities,  might  greatly  exceed  gains, 
which  would  then  be  positive  quantities. 


AN  EXTENSION  OF  THE   IDEA   OF   NUMBER  15 

11.  From  the  nature  of  things,  we  may  treat  positive  and  nega- 
tive quantities  according  to  the  following  Principles  : 

Principle  I.  If  a  positive  and  a  negative  quantity  of  the  same 
kind  are  equal  in  absolute  value^  either  will  destroy  the  effect  of  the 
other  when  both  are  taken  together  oi"  combined  by  addition. 

E.  g.  Items  of  income  and  expense  may  be  regarded  as  being  opposite 
quantities,  and  we  may  call  one  positive  and  the  other  negative ;  for  any 
item  of  expense  reduces  by  just  an  equal  amount  the  effective  income. 

Principle  II.  Positive  quantities  alone  may  be  added  in  any 
order  ;  also  negative  quantities  alone  may  be  added  in  any  order. 

E.  g.  Since  negative  quantities  are  those  which  are  considered  as  tending 
to  diminish  the  effect  of  certain  others  called  positive  quantities,  the  com- 
bined eflfect  of  several  negative  quantities  will  be  a  negative  quantity  which 
is  equal  to  their  sura. 

There  is  no  contradiction  in  speaking  of  adding  negative  quanti- 
ties, for  the  idea  suggested  by  the  terms  positive  and  negative  is  one 
of  nature  or  quality^  not  number  or  amount. 

If  incomes  be  regarded  as  positive,  expenses  must  be  treated  as  negative 
quantities,  and  we  may  add  all  of  our  expenses  and  then  subtract  the  sum 
total  from  our  income  to  determine  our  financial  condition. 

A  single  negative  quantity  may  "oppose"  a  positive  quantity  to  produce 
a  decreased  "value"  indicated  by  subtraction,  while  taken  with  another 
negative  quantity  there  will  be  produced  an  "increased  negative  effect" 
which  would  have  to  be  indicated  by  addition. 

Thus,  as  before,  the  total  expense  results  from  adding  several  expenses. 

Principle  III.     The  resultant  effect  of  several  combined  positive 
and  negative  quantities  is  equal  to  the  numerical  difference  between  ■ 
the  total  positive  and  total  negative  effects^  and  has  the  quality  or 
nature  of  the  greater  total. 

E.g.  The  result  of  combining  expenses  of  $5  and  $10  with  items  of 
income  of  $3,  $2,  $3,  |6,  $4,  $2,  and  $1,  may  be  obtained  by  finding  the 
difference  between  the  total  expense,  $15,  and  the  total  income,  $21.  This 
difference  would  be  a  balance  of  |6  in  favor  of  the  income.  This  balance 
may  be  taken  as  a  positive  quantity. 

Principle  IV.  The  removal  or  subtraction  of  a  positive  quantity 
has  the  same  effect  on  an  expression  in  which  it  occurs  as  the  addition 
of  a  negative  quantity  equal  in  ahsolate  value  to  the  positive  quantity. 


16  FIRST  COURSE  IN  ALGEBRA 

E.  g.  Consider  the  items  of  income  $3,  $2,  and  $4  as  positive  quantities. 
The  effect  produced  on  the  total  income  of  neglecting  or  subtracting  one  of 
these  items,  say  the  amount  of  $4,  may  also  be  produced  by  adding  or  in- 
curring an  expense  of  $4,  since  each  results  in  diminishing  the  effective 
income  by  ^4. 

The  "  not "  taking  of  one  thing,  say  an  item  of  income,  amounts  in  effect 
to  taking  an  item  of  opposite  character  or  quality,  that  is,  an  "  equal "  item 
of  expense. 

Principle  V.  The  removal  or  subtraction  of  a  negative  quantity 
has  the  same  effect  on  an  eo'pression  in  which  it  occms  as  the  intro- 
duction oJ\  or  addition  oJ\  a  positive  quantity  equal  in  absolute  value 
to  the  negative  quantity. 

E.g.  In  order  to  restore  a  given  amount  of  money  to  its  original  value 
after  incurring  an  expense  of  ^4,  it  is  necessary  to  bring  about  an  increase 
of  ^4. 

If,  instead  of  spending  and  then  earning  equal  amounts,  we  neglect  to 
spend,  that  is,  if  we  ttike  away  or  subtract  an  item  of  expense,  our  original 
capital  remains  unaltered. 

Ex.  1.  Chissify  the  changes  in  temperature  from  64°  F.,  to  110°  F.  and 
to  32°  F.,  respectively. 

Since  we  have  an  increase  in  temperature  in  changing  from  64°  F.  to 
110°  F.  and  a  decrease  in  changing  from  64°  F.  to  32°  F.,  we  may  regard 
one  of  these  changes  as  being  positive  and  the  other  negative. 

Thus,  if  the  increase  be  taken  positive  the  decrease  must  be  regarded 
n^ative. 

Exercise  II.    1 

Classification  of  Quantities  as  Being  Either  Positive 
or  Negative 

Classify  the  following  changes  in  temperature  as  being  both  posi- 
tive, both  negative,  or  one  positive  and  the  other  negative  : 

1.  From  60°  F.,  to  100°  F.  and  to  50°  F.  respectively. 

2.  From  68°  F.,  to  90°  F.  and  to  212°  F.  respectively. 

3.  From  76°  F.,  to  0°  F.  and  to  32°  F.  respectively. 

4.  From  102°  F.,  to  40°  F.  and  to  80°  F.  respectively. 

5.  From  0°  F.,  to  17°  F.  below  zero  and  to  5°  F.  below  zero  respectively. 

6.  From  11°  F.  below  zero,  to  21°  F.  below  zero  and  to  15°  F.  below 
zero  respectively. 


AN  EXTENSION   OF  THE  IDEA   OF  NUMBER  17 

7.  From  50°  F.  above  zero,  to  10°  F.  below  zero  and  to  10°  F.  above 
zero  respectively. 

8.  From  6°  F.  above  zero,  to  20°  F.  below  zero  and  to  5°  F.  below  zero 
respectively. 

9.  From  10°  F.  below  zero,  to  18°  F.  below  zero  and  to  9°  F.  below  zero 
respectively. 

10.  From  16°  F.  below  zero,  to  14°  F.  below  zero  and  to  3°  F.  below 
zero  respectively. 

Which  of  the  following  cities  may  be  selected  as  points  of  refer- 
ence in  order  that  the  distances  to  the  remaining  two  may  be 
classified  as  being  both  positive  or  both  negative? 

Ex.  11.    Boston,  Atlanta,  Baltimore. 

Boston  and  Baltimore  are  both  north  of  Atlanta.  Accordingly,  the  dis- 
tances of  these  cities  from  Atlanta  are  both  measured  in  the  same  direction. 
Accordingly,  both  may  be  taken  as  positive  or  both  negative. 

Also,  since  Baltimore  and  Atlanta  are  both  south  of  Boston,  the  distances 
from  Boston  to  Baltimore  and  Atlanta  may  be  taken  as  both  positive  or  both 
negative. 

12.  New  York,  Philadelphia,  Washington,  D.  C. 

13.  New  Orleans,  San  Francisco,  Montreal. 

14.  Boston,  London,  Madrid. 

Which  of  the  following  cities  may  be  selected  as  points  of  reference 
in  order  that  the  distances  to  the  remaining  two  may  be  classified 
as  being  one  positive  and  the  other  negative  1 

15.  London,  Paris,  Rome. 

16.  St.  Petersburg,  Calcutta,  Pekin. 

17.  Boston,  Buffalo,  Chicago. 

Regarding  a  man's  income  as  representing  a  positive  quantity, 
classify  the  following  items  as  positive  or  negative  quantities  wher- 
ever possible  : 

18.  (a)    Money  loaned  to  a  friend. 

(b)  Interest  paid  on  a  mortgage. 

(c)  Interest  on  money  deposited  in  the  bank. 

{d)   Money  drawn  out  of  one  bank  and  deposited  in  another. 
(e)    Money  paid  for  house  rent. 

Regarding  money  on  hand  as  representing  a  positive  value,  classify 
the  following  items,  wherever  possible,  as  positive  or  negative  quan- 
tities with  reference  to  the  depositor  : 

2 


18  FIRST  COURSE  IN  ALGEBRA 

19.  (a)   Money  deposited  in  the  bank. 

(6)    Interest  received  on  money  deposited  in  the  bank. 

(c)  Interest  paid  on  a  mortgage  held  by  the  bank. 

(d)  Money  withdrawn  from  the  bank. 

20.  Classily  the  items  above  with  reference  to  the  bank. 

With  reference  to  the  equator,  classify  the  latitudes  of  the  fol- 
lowing places  as  being  positive  or  negative : 

21.  Mmiich,  Vienna,  Buenos  Ayres,  Quito,  Glasgow,  St.  Louis,  Mel- 
bourne, Zanzibar. 

Since,  starting  at  the  equator,  it  would  be  necessary  to  travel  north  to 
reach  Munich,  Vienna,  Glasgow,  and  St.  Louis,  and  to  travel  in  the  opposite 
direction,  that  is,  south,  to  reach  Buenos  Ayres,  Quito,  Melbourne,  and  Zanzi- 
bar, the  distances  from  the  equator  to  the  places  first  named  may  he  con- 
sidered as  being  all  positive  or  all  negative.  Accordingly,  the  distances 
from  the  equator  to  the  places  last  named  would  be  regarded  as  being  either 
all  negative  or  all  positive,  respectively. 

22.  Tokio,  Jerusalem,  Sidney,  Stockholm,  Honolulu,  Rio  Janeiro,  Cape 
Town,  Tunis. 

With  reference  to  the  meridian  passing  through  Greenwich,  classify 
the  longitudes  of  the  following  places  as  being  positive  or  negative 
quantities: 

23.  Shanghai,  Minneapolis,  Berlin,  Naples,  Dublin,  Ottawa,  Havana. 

Which  of  the  following  dates  must  be  selected  for  reference  in 
order  that  the  changes  in  time  to  the  remaining  dates  may  be  clas- 
sified as  one  positive  and  the  other  negative  ? 

24.  (a)  1492,  1620,  1776. 
(6)  1812,  1861,  1863. 

(c)  44  B.C.,  64  A.D.,  753  B.C. 

(d)  1815,  1066,  1349. 

(e)  490  B.C.,  480  B.C.,  146  A.D. 

The  Invention  of  Negative  Numbers 

12.  Imagine  a  series  of  equal  steps  or  distances  to  be  laid  off  along 
a  straight  line,  unlimited  in  length,  taken  for  convenience  in  a  ver- 
tical position,  as  in  Fig.  1.  Then,  beginning  with  the  lower  end  of 
the  line  and  counting  upward  we  will  number  the  points  of  division, 
using  the  primary  numerals  1,  2,  3,  4,  5,  etc. 


AN  EXTENSION   OF   THE   IDEA  OF   NUMBER  19 

If  we  regard  motion  upward,  or  counting  upward  along  the  "  car- 
rier "  line,  as  being  in  a  positive  direction,  then  motion  downward, 
or  counting  downward,  must  be  regarded  as  being  negative. 

13.  Fixing  our  attention  on  any  particular  number  we  may 
find  a  larger  number  by  counting  upward,  and,  except  in  the 
case  of  1,  a  smaller  number  by  counting  downward.  Any 
number  will  be  relatively  positive  with  regard  to  another  if 
it  be  situated  above  it  in  position,  and  relatively  negative 
to  it,  if  it  be  necessary  to  count  downward  from  the  other  to 
find  it. 

E.  g.  Relatively  to  8,  10  is  positive,  while  all  smaller  numbers, 
as  5,  4,  3,  etc.,  are  negative,  since  we  should  have  to  count  down- 
ward from  8  to  reach  5,  4,  3,  etc. 

14.  Observe  that  it  is  possible  for  us  to  count  upward  for  any 
number  of  spaces^  starting  anywhere  in  the  series  1,  2,  3,  4,  etc.,  but 
it  is  not  possible  to  count  downward  for  any  number  of  spaces.  This 
is  because  we  must  always  stop  when  we  reach  the  lowest  point  of 
the  line,  since  there  are  no  numbers  below  it  to  count. 

Furthermore,  we  cannot  count  downward  at  all  if  we  start  at  the 
lowest  point,  1,  for  we  have  reached  the  end  of  our  line,  and  at  the 
same  time  the  "  lower  "  end  of  our  series  of  integral  primary  numbers. 

15.  There  is  nothing  unreasonable  in  imagining  our  line  to  be 
now  extended  downward,  carrying  our  series  of  steps  downward 
indefinitely. 

In  order  to  distinguish  these  newly  added  downward  steps  from 
those  above  our  starting  point,  which  we  will  take  as  0,  we  may 
designate  them  by  saying  one  below  zero^  two  below  zero^  three  below 
zero,  etc. 

16.  Since  starting  from  0,  it  is  necessary  to  count  in  opposite 
directions  along  our  "carrier"  line  to  reach  numbers  having  the 
same  number  name  (as,  for  example,  "four  above"  zero  and  "four 
below"),  we  may  distinguish  the  two  sets  of  numbers  by  calling 
them,  relatively  to  0,  one  set  positive  and  the  other  negative. 

17.  We  will  call  the  numbers  "  above  "  zero  positive  one,  positive 
two,  positive  three,  etc.  (written  "^1,  "^2,  '^'S,  etc.),  and  those  "  below  " 
zero  negative  OTie,  negative  two,  negative  three,  etc.  (written  ~1,  ~2,  ~3, 
etc.). 


20 


FIRST  COURSE   IN   ALGEBRA 


In  this  and  the  next  three  chapters  we  shall  indicate  the  "  quality  " 
of  a  number  as  being  positive  or  negative  by  writing  before  it  a  small 
"  quality  "  sign  "^  or  ~  When  so  used  these  signs  are  read  j^ositive 
and  negative  respectively,  and  are  called  signs  of  quality.  They 
will,  by  their  size  and  position,  be  easily  distinguished  from  the  larger 
signs  of  operation  for  addition  or  subtraction,  +  and  — ,  which  are 
read  plus  and  minus  respectively. 

18.    We  have  applied  the  Principle  of  No  Ex- 
I  ception  to   our  notion   of  counting,  and   have 

"stretched  out,"  or  "extended"  our  number 
series  to  allow  of  the  idea  of  counting  "back- 
ward "  or  "  downward  "  beyond  zero. 

The  important  point  to  be  understood  is  that 
the  positive  numbers  """l,  "^2,  +3,  "^4,  etc.,  repre- 
sented on  Fig.  2  beside  the  "black  circles"  or 
dots,  should  be  regarded  as  having  arisen  as  the 
result  of  the  actual  counting  of  objects. 

On  the  other  hand,  the  negative  numbers, 
~1,  ~2,  ~3,  ~4,  etc.,  represented  on  the  figure 
beside  the  small  rings,  or  "white  circles,"  were 
invented  simply  to  serve  our  convenience,  in  or- 
der that  we  might  represent,  or  imagine,  count- 
ing "  downward  "  or  "  backward  "  below  zero. 

These  negative  numbers  may  be  regarded  as 
being  "artificial"  numbers,  and  as  being  simply 
the  invention  of  mathematicians  to  serve  as  con- 
venient means  for  simplifying  work  and  inter- 
preting results. 
I 

I  19.    Letters  were  first  used  to  represent  negative 

j'jg    2.  numbers  by  Descartes  in  the  first  half  of  the  seven- 

teenth century.  In  a  book  published  in  1545  by  an 
earlier  writer,  Cardan,  they  were  called  "  numeri  ficti,"  or  imaginary  num- 
bers, in  contradistinction  to  "  numeri  veri,"  or  real  numbers. 

At  the  present  time  the  term  "  imaginary "  number  is  applied  to  still 
another  form  of  "invented"  number.     (See  Chapter  XXI.) 

20.'  By  the  absolute  value  of  a  number  or  letter  is  meant  its 
value  without  regard  to  its  quality  as  being  positive  or  negative. 
(See  also  §  7.) 


£        ^ 

►    "6 

V 

■^5 

J        * 

+4 

T        ' 

I        i 

►    "3 

S        ^ 

n 

0 

^    -^1 

p      ' 

)    ^^ 

-0     ^ 

)        ^ 

-1   ^ 

E 

-2    ^ 

G 

-3    ^ 

1         ^ 

)        T 

-4 

>        / 

-5     ^ 

^ 

-6    ^ 

1^        E 

AN   EXTENSION   OF  THE   IDEA   OF  NUMBER  21 

The  absolute  value  of  a  number  or  letter  is  indicated  by  writing 
it  between  two  upright  bars,  |    |  . 

E.g.    \a\  means  the  absolute  value  of  some  number  a. 
We  may  write  l+a|  =  |~«|-     This  is  read  :  "The  absolute  value  of  posi- 
tive a  is  the  same  as  the  absolute  value  of  negative  a." 

By  the  arithmetic  value  of  a  number  is  meant  its  absolute  value. 
E.  g.    The  arithmetic  value  of  ""4  is  4. 

21.  Just  as  in  arithmetic  we  regard  any  whole  number  as  being 
a  repetition  of  the  primary  unit  1  a  certain  number  of  times  (for 
example  5  =  1  +  1  +  1  +  1  +  1),  so  in  algebra  we  regard  positive 
and  negative  numbers  as  being  repetitions  of  the  quality  units 
+1  and  "1. 

Of  these  quality  units,  "''1  is  taken  as  the  primary  unit. 

A  positive  number  is  the  sum  of  two  or  more  positive  units,  and 
a  negative  number  is  the  sum  of  two  or  more  negative  units. 

Positive  numbers  and  negative  numbers  taken  together  are  called 
algebraic  numbers. 

E.g.  ■♦"5  denotes  five  times  +1,  or  five  positive  units.  Hence  we  may 
write  +5  =  +  +1  +  +1+  +1+  +1  +  +1- 

-5  denotes  five  times  -1,  or  five  negative  units.  Hence  we  may  write 
-5  =  +  -l+-I+-i+-l+-l. 

22.  The  sign  of  continuation ,  read  and  so  on,  is  used 

to  indicate  that  the  expression  as  written  may  be  extended. 

Thus,  the  expression  1  +  2  +  3  + ,  may  be  extended,  if 

desired,  as  for  example, 

1  +  2  +  3  +  4  +  5  +  6  +  7  +  8  + 

The  sign  of  continuation  may  also  be  used  to  indicate  that  certain 
parts  of  an  expression  have  been  omitted  for  convenience. 

Thus,  in  the  expression  2  +  4  +  6  + +96  +  98  +  100, 

the  sum  of  the  numbers  from  8  to  94  inclusive  has  been  omitted. 

23.  The  symbol  oo  which  is  read  "  infiinity,  "  is  used  to  represent  a 
number  which  is  greater  than  any  assignable  number,  however  great. 

We  may  symbolize  the  whole  series  of  algebraic  number  by 
~^ ,  -3,  -2,  -1,  ±0,  +1,  +2,  +3, ,  +00,  the  order  of 


22  FIRST  COURSE  IN  ALGEBRA 

ascending  magnitude  being  from  the  left  to  the  right.  The  double 
sign  *,  read  "  positive  or  negative,  "  is  placed  before  the  zero  to  in- 
dicate its  exceptional  property,  namely,  that  "•■()  =  ~0,  We  may  re- 
gard zero  as  belonging  to  both  the  positive  and  negative  parts  of  the 
series.     It  is  all  that  these  parts  have  in  common. 


LAWS  FOR  ADDITION  AND  SUBTRACTION  23 


CHAPTER  III 

FUNDAMENTAL  LAWS  OF  ALGEBRA   FOR  THE  ADDITION  AND 
SUBTRACTION  OF  POSITIVE  AND  NEGATIVE  NUMBERS 

1.  Since  in  order  to  obtain  correct  results  we  must  work  with 
numbers  according  to  certain  rules,  it  is  common  to  speak  of  them 
as  obeying  laws.  What  we  mean,  in  reality,  is  that,  unless  we  obey 
certain  rules  or  laws  in  performing  our  calculations,  we  cannot  depend 
upon  the  accuracy  of  our  results. 

2.  It  follows  from  the  definitions  of  the  number  symbols  of  arith- 
metic that  since  2  =  1  +  1,  and  3  =  1  +  1  +  1,  that  +  3  +  2  = 
^-  2  +  3 ;  and  in  the  same  way  we  may  show  that  +  3  +  4  +  7  = 
+  4  +  3  +  7  =  +  7  +  ^^+4,  etc.,  and  in  general  if  a,  6,  c,  etc.,  repre- 
sent any  whole  numbers  in  arithmetic,  a  +  b  +  c  =  a  +  c  +  b  = 
b  +  c  +  a  =  etc. 

Hence  in  a  chain  of  additions  the  result  is  independent  of  the  order 
in  which  the  additions  are  performed. 

This  is  called  the  Law  of  Commutation  for  addition  in  arith- 
metic, and  it  remains  for  us  to  show  that  it  can  be  applied  to 
expressions  which  contain  both  positive  and  negative  numbers,  that 
is,  algebraic  numbers. 

3.  As  symbols  of  grouping  we  have  the  parentheses  (  ),  brackets 
[  ],  braces  {  },  and  vinculum  which  is  sometimes  written 
vertically  |. 

These  symbols  indicate  in  each  case  that  the  expression  included 
is  to  be  operated  with  as  a  whole.     (See  also  Chapter  IV.) 

E.  g.  (3  +  5)  means  that  the  sum  of  3  and  5  is  to  be  used  as  a  whole  in 
un  operation,  that  is,  as  8,  so  that  an  expression  such  as  2  x  (3  +  5)  is  to  be 
understood  as  meaning  two  times  8,  or  16  ;  while  the  expression  2x3+5 
means  two  times  3,  or  6,  increased  hy  5,  — that  is,  11,  not  16. 

Again,  (4  +  2)(8  —  1)  means  4  +  2,  that  is,  6,  multiplied  by  8  —  I,  or  *7, 
that  is,  6  X  7,  or  42  ;  while  4  +  2x8  —  1  means  4  increased  hy  2  tim£s  8, 
or  16,  and  this  sum  diminished  hy  1.     Hence  the  result  is  19. 


24  FIRST  COURSE  IN  ALGEBRA 

4.  It  is  a  matter  of  experience  with  us  in  arithmetic  that  in  a 
chain  of  additions  tJw  sum  total  is  not  affected  by  combining  two  or 
more  parts  of  the  sum. 

This  i§  called  the  Law  of  Association  for  addition. 

E.  g.    3  +  5  +  6  is  equal  to  (3  +  5)  +  6,  that  is,  14  is  equal  to  8  +  6. 
Again,  the  original  expression  is  equal  to  3  +  (5  -|-  6),  that  is,  14  is  also 
equal  to  3+  11. 

If  rtr,  b,  c,  etc.,  represent  any  whole  numbers  in  arithmetic,  we 
may  state  the  Law  of  Association  in  symbols,  by  writing 

a  +  6  +  c  =  (a  +  &)  +  c  =  «  +  (6  +  c). 

5.  The  foundations  of  all  mathematical  knowledge  must  be  laid 
in  definitions. 

A  definition  is  an  explanation  of  what  is  meant  by  any  word  or 
phrase,  and  must  be  given  in  terms  of  things  other  than  those  con- 
sidered. It  is  essential  to  a  complete  definition  that  it  distinguish 
perfectly  the  thing  defined  from  everything  else. 

In  mathematics  the  principal  terms  may  be  so  defined  as  to  leave 
not  the  slightest  question  respecting  their  meaning. 

6.  There  are  comparatively  few  mathematical  truths  or  theorems 
which  are  self-evident.  The  majority  require  to  be  proved  by  a 
chain  or  course  of  reasoning.  The  course  of  reasoning  by  which  the 
truth  of  a  statement  is  established  is  called  a  demonstration  or 
proof. 

As  symbols  of  deduction  we  have  in  algebra,  as  in  arithmetic, 
.  *.  meaning  ther^ore,  and  *. '  meaning  siiice  or  because. 

Steps  on  a  Line 

7.  We  shall  speak  of  different  distances  measured  on  the  "  carrier  " 
line  which  "  carries  "  our  collection  of  extended  number,  ""3,  ~2,  ~1, 
^0,  "^1,  "*"2,  "^3,  etc.,  as  steps.  We  shall  say  that  a  step  is  2>ositive 
if  in  taking  it  we  step  "up,"  that  is,  in  the  direction  of  the  increas- 
ing primary  numbers,  1,  2,  3,  4,  etc.  We  shall  call  it  negative  \i  we 
step  "down,"  that  is,  in  the  direction  +6,  +5,  +4,  +3,  +2,  +1,  or  "1, 
-2,  "3,  -4,  -5,  etc.     (See  Fig.  1.) 

8.  We  shajl  understand  that  two  steps  taken  anywhere  on  the 


LAWS   FOR   ADDITION   AND   SUBTRACTION  25 

line  are  equal,  when  their  lengths  on  the  line  are  equal,  and  when 
they  are  taken  in  the  same  direction  or  sense. 

E.  g.  The  step  from  +5  to  +8  is  equal  to  the  one  from  +8  to  +11 ;  or 
from  +12  to  +15  ;  or  from  +20  to  +23 ;  etc.  It  is  also  equal  to  the  step 
from  ~7  to  -4;  from  —8  to  ~5;  and  these  are  all  to  be  considered  as  positive 
steps. 

As  negative  steps,  each  having  a  length  of  four  units,  we  may  select  the 
one  from  +10  "  down"  to  +6;  from  +4  to  +0;  from  +2  to  "2;  from  "0  to 
-4 ;  from  ~5  to  ~9  ;  etc.  , 

9.  If  we  move  along  the  "  carrier  "  line,  taking  successively  two 
steps,  either  both  in  the  same  direction,  or  the  first  in  one  direction, 
then  turning  around,  take  the  other  in  the  opposite  direction,  we 
speak  of  the  process  as  effecting  the  composition  of  the  two  steps. 

10.  Tivo  separate  steps  mai/,  by  composition,  be  combined  into  a 
single  step. 

E.  g.  Two  separate  steps  of  3  and  5  units  respective!}'-,  may  by  composi- 
tion be  combined  into  a  single  step  of  8  units  in  length,  and  the  fact  that 
both  are  positive  or  both  negative  will  not  affect  the  result. 

11.  The  single  step  which  may  be  t^ken  in  place  of  two  or  more 
others  taken  successively,  is  called  the  resultant  step.  The  re- 
sultant step  may  be  defined  as  the  single  step  which  may  be  taken 
to  produce  the  same  final  change  of  position  as  that  produced  by 
taking  several  others  successively. 

If  we  travel  along  the  line  taking  different  steps  successively, 
observe  that  the  resultant  step  is  measured  from  the  beginning  to 
the  end  of  the  journey.  It  is  sometimes  shorter,  but  never  longer  than 
the  entire  path  passed  over. 

E.  g.  If  the  steps  +7,  ~5,  and  +1  were  taken  successively,  starting  from 
any  point  of  the  line,  as  say  +3,  we  should  arrive  first  at  +10  on  the  line; 
then  returning  by  tlie  step  ~5,  stop  at  +5,  and  tlie  end  of  our  journey  would 
be  one  unit  "  higher,"  or  at  +6.  Since  the  distance  from  the  starting  point, 
+3,  to  the  stopping  point,  +6,  is  three  units,  taken  in  an  "  upward  "  or  pos- 
itive direction,  we  say  that  the  resultant  of  the  steps  +7,  ~"5,  +1,  is  the 
single  step  +3. 

Similarly,  the  resultant  of  +6,  ~4,  "5,  taken  successively,  would  be  a 
single  negative  step  of  three  units,  or  in  symbols,  the  step  ~3. 


26  FIRST  COURSE   IN   ALGEBRA 

12.  The  following  Principles  will  be  seen  to  apply  to  steps  tali'en 
along  a  line. 

Principle  I.  Tlie  resultant  of  two  steps  of  the  same  kind^  that 
is,  of  two  steps  taken  in  the  same  directionyis  a  single  step  of  the  same 
kind.  Its  length  is  equal  to  the  sum  of  the  lengths  of  the  single  steps 
composing  it. 

E.  g.    The  resultant  of  the  separate  steps  +4,  +3,  +5,  is  the  single  step  +12. 

Principle  II.  The  resultant  of  two  steps  taken  successively  in 
opposite  directions  is  a  single  step  whose  length  is  the  difference  he- 
tiveen  the  lengths  of  the  separate  steps  composing  it.  This  is  a  posi- 
tive or  a  negative  step,  according  as  the  greater  o?ie  entering  into  it 
is  positive  or  fiegative. 

E.  g.  The  resultant  of  the  two  steps  +12  and  "5  is  a  step  of  12  minus  5, 
that  is,  7  units  in  length,  and  since  the  greater  of  the  two  steps  entering  into 
it,  +12,  is  positive,  we  must  have  as  a  resultant  +7. 

Principle  III.  The  resultant  of  two  or  more  steps  does  not  depend 
upon  the  order  iti  which  the  steps  are  taken. 

E.  g.  The  resultant  of  steps  +12,  "8,  +3  and  -2,  taken  in  any  order,  is 
the  single  step  +5. 

Principle  IV.  The  resultant  of  any  number  of  steps  is  a  single 
step  whose  length  is  the  difference  between  the  sum  of  the  lengths  of  the 
positive  steps  and  the  sum  of  the  lengths  of  the  negative  steps.  This 
resultant  is  positive  or  negative  according  as  the  sum  total  of  the  pos- 
itive steps  or  of  the  negative  steps  is  the  greater. 

E.  g.  The  resultant  of  the  separate  steps  +10,  +5,  ~3,  ~6,  +1  may  be 
found  by  taking  the  resultant  of  the  total  positive  step  +(10  +  5  +  1)  or 
+16,  and  the  total  negative  step  -(3  +  6)  or -9.  The  difference  between 
16  and  9  is  7,  and  since  the  greater  of  the  two  resultant  steps,  +16  and  ~9, 
is  positive,  the  resultant  step  is  positive,  and  we  have  as  a  resultant  +7. 

The  resultant  of  the  successive  steps  +11,  "8,  ~6,  +2,  "4,  is  found  by 
taking  the  resultant  of  the  total  positive  step  +13,  and  the  total  negative 
step  ~18,  which  is  ~5. 

13.  The  taking  of  steps  along  the  number  series  suggests  the 
operations  of  arithmetic  addition  and  subtraction. 

Taking  steps  "upward,"  that  is,  in  the  direction  of  increasing 


LAWS   FOR   ADDITION  AND  SUBTRACTION  27 

primary  numbers,  5,  6,  7,  8,  etc.,  suggests  arithmetic  addition. 
Taking  steps  "downward,"  that  is,  in  the  direction  of  decreasing 
primary  numbers,  8,  7,  6,  5,  etc.,  suggests  arithmetic  subtraction. 

14.  We  will  now  find  an  interpretation  for  combinations 
of  si^ns  of  operation  and  signs  of  quality  such  as  +("^«), 
-(""«),  +("«)  and  -(-«). 

In  order  to  do  this,  we  will  now  interpret  our  signs  of  operation 
to  mean  :  plus  +,  take,  that  is,  to  include  with  other  steps,  and 
minus  — ,  take  away,  that  is,  remove  from  an  expression  or  neglect 
in  connection  with  other  steps. 

These  ideas  correspond  to  arithmetic  addition,  or  "taking  one 
number  with  another,"  and  arithmetic  subtraction,  which  is  "tak- 
ing one  number  away  from  another,"  or  neglecting  it  from  a  sum. 

15.  The  effect  on  a  mm  total  of  taking  away  a  positive  step  is  the 
same  as  that  of  performing  a  negative  step  of  equal  numerical  value 
or  length. 


E.  g.    From  the  step  "^5  take  away  the  step  +3.  ^^>^ 

We  have  +5  -  +3  =  +(5  -  3)  =  +2. 

Also,  from  another  point  of  view, 

+5  +  -3  =  +2. 

We  obtain  the  same  result  as  before,  using  this  time 
a  negative  step  in  an  additive  sense,  instead  of  using,  as 
in  the  first  place,  a  positive  step  in  a  subtractive  sense. 
(See  Fig.  2.)  ^^  J^ 


*5 
M 
+  3 

'0 


16.   A  negative  step  produces  a  decrease  of  value  among  positive 
numbers. 

Hence,  taking  away  a  decrease  amounts  to  making  an  increase. 
Therefore,  -(~a)  =  +{-^a). 

Hence,  from  the  illustration  above  we  may  draw  the  following 
conclusions  : 

(i.)    +  (+«)  =  +  (+«) ;  (iii.)  -  (-a)  =  +  (+o)  ; 

(ii.)   -  (+«)  =  +  (-a)  ;  (iv.)   +  (-«)  =  +  (-a). 

It  will  be  seen  that  in  the  identities  above,  the  symbol  of  opera- 
tion +  occurs  in  every  case  on  the  right  of  the  identity  sign ;  that 


28  FIRST  COURSE   IN  ALGEBRA 

is,  we  have  transformed  our  additions  and  subtractions  on  the  left 
into  additions  on  the  right. 

Furthermore,  wherever  the  symbol  of  subtraction  on  the  left,  as 
in  (ii.)  and  (iii.),  has  been  changed  into  that  of  addition  on  the 
right,  the  sign  of  quality  has  also  been  reversed. 

17.  We  may  now  state  the  Law  of  Sig^us  for  Addition  and 
Subtraction. 

Principle:  Additions  may  he  substituted  for  subtractions,  pro- 
vided that  the  signs  of  quality  of  the  numbers  subtracted  be  reversed. 

Ex.  1.    From  the  step  +8  subtract  the  step  +3. 

Indicating  the  subtraction   by  writing  +8  —  +3,  we  may  transform  the 
expression  so  that  instead  of  a  subtraction  we  shall  have  an  indicated  addi- 
tion.    By  changing  the  symbol  of  operation  before  3  from  —  to  +,  and  at 
the  same  time  reversing  the  sign  of  (quality  and  writing  -3,  we  have 
+8- +3  =  +8 +  -3  =  +5. 
Ex.  2.    From  the  step  +10  subtract  the  step  -4. 
Indicating  the  subtraction  as  before,  we  have 

+10  -  -4. 
Substituting  an  addition  for  the  subtraction,  and  reversing  at  the  same 
time  the  sign  of  quality,  we  have 

+10  -  -4  =  +10  +  +4  =  +14. 
Ex.  3.    From  the  step  —8  subtract  the  step  +2.     As  before  we  may  write 

-8  _  +2  =  -8  +  -2  =  -10. 
Ex.  4.    From  the  step    9  subtract  the  step  "6. 
We  have  -9  -  "6  =  "9  +  +6  =  "3. 

18.  If  instead  of  the  word  "  step  "  we  substitute  the  word  "num- 
ber," or  the  more  general  word  "quantity,"  we  may  regard  the  prin- 
ciples of  this  chapter  as  being  principles  governing  operations  with 
positive  and  negative  numbers  or  quantities. 

19.  Whenever  reference  is  made  to  signs  in  algebra  it  is  to  be 
understood,  unless  the  contrary  is  stated,  that  the  +  or  —  signs 
are  meant.  Thus,  when  we  speak  of  the  sign  of  a  quantity  we  shall 
mean  the  +  or  —  sign  which  is  prefixed  to  it  or  its  numerical  coeffi- 
cient, not  the  X  or  -f-  signs  which  may  be  associated  with  it. 

When  we  are  directed  to  change  the  signs  of  an  expression  we 
shall  understand  that  we  are  to  change  the  +  or  —  signs  before 
every  term  into  —  or  +,  respectively. 


POSITIVE  AND  NEGATIVE  NUMBERS  29 


CHAPTER  IV 

ADDITION  AND   SUBTRACTION  OF  POSITIVE  AND 
NEGATIVE  NUMBERS 

Symbols  of  Grouping. 

1.  In  order  to  denote  that  an  algebraic  expression  is  to  be  treated 
as  a  whole,  in  a  calculation,  it  is  enclosed  within  parentheses  or 
other  symbols  of  grouping.     (See  Chapter  III,  §  3.) 

E.  g.  (2  +  5  +  7)  is  to  be  regarded  not  as  three  different  numbers,  2,  5, 
and  7,  but  as  the  number  obtained  by  adding  5  and  7  to  2,  which  is  14. 

Also,  (5  +  2)  X  (3  +  8)  means  7  multiplied  by  11,  that  is,  77;  while 
5  +  2x3  +  8  means  5  increased  by  6,  increased  by  8,  that  is,  19. 

2.  Symbols  of  grouping  may  be  of  different  kinds ;  thus, 
parentheses  ( ),  braces  {},  brackets  [  ],  etc.  The  effect  in 
each  case  is  the  same,  namely,  to  call  our  attention  to  the  fact  that 
whatever  is  enclosed  in  them  is  to  be  treated  or  regarded  as  a  whole. 

Occasionally  it  is  convenient  to  use  instead  of  parentheses  a  line 
called  a  vinculum,  drawn  over  an  expression. 


Thus,  7  +  5  —  2  is  equivalent  to  7  +  (5  —  2).     This  notation  will  be 

used  commonly  in  connection  with  fractions  and  radical  or  root  signs. 

2  2 

E.  g.  means  2  divided  by  (3  +  4),  that  is,  -. 


\/\\  +  5  means  the  square  root  of  (11  +  5),  that  is,  the  square  root  of 
16,  which  is  4. 

3.  Expressions  containing  groups  of  terms  enclosed  in  parentheses 
may  be  treated  as  follows  : 

Ex.  1.    Consider  the  expression  25  —(15  +  3). 

This  expression  means  that  the  sum  of  15  and  3,  which  is  18,  is  to  be 
taken  from  25  to  produce  the  remainder  7. 

To  subtract  18  from  25  in  a  single  operation  amounts  to  performing  the 
two  separate  operations  of  first  subtracting  15  from  25,  and  then  decreasing 
the  remainder  by  3.     The  final  result  is  7, 


30  FIRST  COURSE  IN   ALGEBRA 

Hence  ?5  -  (15  +  3)  =  25  -  15  -  3  ^  7. 

Accordingly  -  (15  +  3)  =  -  15  -  3. 

Ex.  2.   Consider  the  expression  12  —  (9  -  2). 

This  expression  means  that  2  is  to  be  first  subtracted  from  9  to  produce 
7,  which  is  then  to  be  taken  from  12,  leaving  5  as  a  final  result,  that  is  : 
12  -  (9  -  2)  =  5. 

If  we  had  first  subtracted  9  we  would  have  diminished  12  by  a  number 
too  great  by  2. 

Hence  it  would  have  been  necessary  to  increase  this  result  by  2. 

Consequently,  we  would  have  obtained  the  same  final  result  as  before 
by  performing  the  following  operations : 

12-9  +  2  =  5. 

Hence  it  appears  that  to  subtract  (9  —  2),  or  7,  in  one  operation,  amounts 
to  first  subtracting  9  and  than  adding  2  in  two  separate  operations,  that  is: 

-  (9  -  2)  =  -  9  +  2. 

4.  The  examples  above  are  illustrations  of  the  General  Principle 
that  ire  may  remove  parentheses  preceded  hy  the  sign  of  subtraction 
— ,  provided  we  at  the  same  time  change  the  signs  of  operation  of  the 
numbers  removed ^  from  +  to  —  or  from  —  to  +. 

5.  In  order  to  prove  a  theorem  true  for  any  algebraic  expression, 
it  is  only  necessary  to  prove  it  for  an  expression  containing  letters 
upon  whose  values  no  restrictions  are  placed. 

6.  We  will  now  proceed  to  establish  the  general  principles  govern- 
ing the  removal  or  insertion  of  parentheses  preceded  by  the  signs  of 
operation  +  or  —  in  chains  of  additions  and  subtractions  for  arith- 
metic numbers.  We  will  then  extend  these  principles  to  include 
positive  and  negative  numbers,  that  is,  algebraic  numbers. 

Representing  any  arithmetic  values  by  letters  a,  b,  c,  at  first  regarding 
a'>-  b  >  c,  we  will  deduce  the  laws  for  the  insertion  and  removal  of  paren- 
theses in  a  chain  of  additions  and  subtractions. 

a-\-  (-\-b  —  c)  means  that  we  are  to  first  diminish  b  by  c,  and  then  to 
add  the  result  to  a.  The  parentheses  about  the  binomial  (+  6  —  c)  indicate 
that  we  are  to  treat  the  expression  included  as  a  whole,  and  the  +  sign  before 
the  parentheses  indicates  that  we  are  to  use  the  result  of  the  operation 
+  h  —  c  to  increase  a. 

Hence,  a+(-\-b  —  c)=a  +  b  —  c. 

Consider  now,  a  —  (+  b  —  c). 


POSITIVE  AND   NEGATIVE   NUMBERS  31 

The  inclusion  of  +  6  —  c  in  parentheses  means  that  we  are  first  to  sub- 
tract c  from  h  and  then  to  take  the  result  from  a. 

If  we  were  first  to  take  h  from  a,  that  is,  to  find  the  difference  a  —h,  we 
would  take  away  too  much  from  a  by  the  quantity  c.  Hence,  we  must 
increase  the  remainder  by  a  value  equal  to  c ;  that  is,  we  must  add  c  to  the 
result. 

Therefore,  a  —  (-{- h  —  c)  =  a —  b  -{-  c. 

7.  Principle  I. 

(i.)  Parentheses  preceded  by  a  ■\-  sign  may  he  removed  ivithout 
altering  the  signs  of  opei^ation  of  the  separate  numbers  remaved. 

(ii.)  Parentheses  preceded  by  a  —  sign  may  be  removed,  pi'oviding 
the  signs  of  operation  preceding  all  of  the  numbers  removed  be  changed 
from  +  to  —,  m\f7'om  —  to  +. 

8.  Since  the  proof  of  any  identity  establishes  the  truth  of  its 
converse,  we  may  state 

Principle  II. 

(i.)  A  chain  of  additions  and  subtractions  may  be  enclosed  within 
parentheses  preceded  by  the  sign  of  operation  fm'  addition  +  without 
making  any  alteration  in  the  signs  of  operation  of  the  numbers 
enclosed.     (Associative  Law  for  addition.     See  Chapter  III.  §  4.) 

(ii.)  A  chain  of  additions  and  subtractions  may  be  enclosed  within 
parentheses  preceded  by  the  sign  of  operation  for  subtraction  —  provid- 
ing the  signs  of  operation  of  all  of  the  numbers  introduced  be  changed 
from  -\-  to  — ,  or  from  —  to  -\-. 

9.  Since  the  reasoning  above  does  not  depend  at  all  upon  the 
lengths  of  the  chains  of  additions  and  subtractions  removed  or 
inserted,  the  principles  hold  for  any  chains  of  additions  and  sub- 
tractions. 

E.  g.  Inclusion  Tritliin  Parentheses 

Preceded  by  Sigrn  +  Preceded  by  Si^n  — 

a  —  h-\-c  —  d—e=a  —  })-\-{c  —  d  —  e)  a—h-\-c  —  d—e=a  —  h-\-c—{(l+e) 

or  =(rt— i+f)— cZ— e.  or  =a—{h—c+d-\-e) 

or  =a—(h—c-{-d)—e. 

10.  When  numbers  are  included  within  several  sets  of  parentheses, 
one  set  within  another,  the  student  will  find  it  convenient  in  certain 


32  FIRST  COURSE   IN  ALGEBRA 

cases  to  begin  by  removing  the  innermost  parentheses  first.     In 
other  cases  it  will  be  better  to  work  with  the  outer  parentheses  first. 
11.   Any  change  in  the  form  of  an  expression  tending  to  lessen 
the  number  of  indicated  operations  is  called  a  reduction. 

Ex.  3.    Reduce  7  -  {  4  +  [  2  -  (8  -  5)  ]  }. 

Method  I  Method  II 

Removing  inner  parentheses  fii-st.  Removing  outer  parentheses  first. 

7-{-H{2-(8-5)]l=7-{4+[2-8+5]l  7-{4+[2-(8-5)]}=7-4-[2-(8-5)] 
=7-{  4+2-8+5}  =7_4_2+(8-5) 

=  7-4-2+8-5  =7_4_2+8-5 

=4.  =  4. 

Exercise  IV.     1 
Find  the  numerical  values  of  the  following  expressions: 


1. 

5+(4-2)+3-(6-3).     5.    5 -(2  + 6-3)  +  7-8-4. 

2. 

10-(9-8)-(7-6).        6.    6- {2 +  (10-6+1)}. 

3. 

(6-4)  +  4-4-(6-4).    7.    11  -  [8  -  (10  -  9  -  6)]. 

4. 

12-(2-4-2)+(2+4+2).      8.    12  -  {9  -  [4  +  (2  -  6)]  +  2} 

9.    3  +  [4  -  (5  -  6  -  5)  +  4]  -  3. 

10.    [15  +  (9  -  6  -  3)]  -  [15  -  (9  +  6  -  3)]. 

11.    20  -  {(20  -  5)  -  [20  -  10  +  (20  -  15)]}. 

12.    18 -[15 -9 -4 -(11-2)-!]. 

13.    11  -  [6  +  8  —  5  —  (12  -  7)  -  8  -  5]. 

14.    [21  -  (7  -  8  -  5)]  -  [21  +  (7  -  5  +  8)]. 

15.    24-  {9-(10-6)-[24-17  +  5-(6-4)]}. 

I.    Addition 

12.  Since  positive  and  negative  numbers  enter  into  algebra,  we 
must  so  extend  our  idea  of  addition  as  to  be  able  to  admit  of  uniting 
positive  and  negative  numbers  in  one  "  sum." 

13.  As  in  arithmetic,  numbers  which  enter  into  a  sum  are  called 
summands. 

14.  Subtraction  is  the  operation  by  which,  when  the  sum  of  two 


POSITIVE  AND  NEGATIVE  NUMBERS  33 

expressions  is  known  and  one  of  them  is  given,  the  other  may  be 
found. 

Subtraction  may  be  regarded  as  the  operation  which  is  the  inverse , 
of  addition,  or  the  process  which  "  undoes  "  addition. 

15.  The  known  sum  is  called  the  minuend,  the  given  expression 
to  be  subtracted  the  subtrahend,  and  the  expression  or  number  to 
be  found  is  called  the  difference  or  remainder. 

The  terms  minuend  and  subtrahend  are  used  in  algebra  in  the 
same  sense  as  in  arithmetic.  In  arithmetic  addition  always  pro- 
duces an  increase,  subtraction  a  decrease. 

16.  In  algebra,  addition,  sum,  and  difference  have  each  a  more 
extended  meaning.  On  account  of  the  introduction  of  the  idea  of 
positive  and  negative  numbers,  an  "  addition  "  may  produce  either 
an  increase  or  a  decrease  in  numerical  value ;  a  "  subtraction  "  may 
produce  either  a  decrease  or  an  increase  in  numerical  value. 

17.  Addition  suggests  taking  a  step  forward ;  subtraction  suggests 
taking  a  step  backward.     (See  Chapter  III.    §§  7,  13.) 

E.  g.  In  5  +  3  =  8,  the  second  term,  +  3,  may  be  regarded  as  the  step 
"  forward  "  from  5  to  8,  while  in  5  —  3  =  2,  —  3  may  be  regarded  as  the 
step  "  backward  "  from  5  to  2. 

18.  In  the  addition  of  algebraic  numbers  the  two  following  con- 
ditions may  arise  : 

(a.)  The  numbers  to  be  added  may  both  have  the  same  quality, 
that  is  be 

Both  Positive  or  Both  Negative. 

Ex.  1.  To  +4  add  +6. 

Here  both  sunmiands  are  positive  numbers.  Consequently  4  positive 
units,  when  united  by  addition  with  6  positive  units,  will  produce  a  total 
of  (4  +  6)  positive  units,  or  10  positive  units,  which  may  be  expressed  as 
follows: 

+4  +  +6  =  +(4  +  6)  =  +10. 

Ex.  2.   To  -5  add  "7. 

Here  both  numbers  are  negative.  By  addition,  5  negative  units  taken 
with  7  negative  units  produce  a  combined  result  of  (5  +  7)  negative  units  ; 
that  is,  in  symbols  : 

-5  4-  -7  =  -(5  +  7)  =  -12. 
3 


34  FIRST  COURSE  IN  ALGEBRA 

19.   By  referring  to  our  scale  of  extended  number  it  may  be  seen  that  to 
take  a  step  of  4  positive  units,  and  then  immediately 
I  another  step  of  6  positive  units,  amounts  to  taking 

in  all  a  single  step  of  10  positive  units.     (See  Ex.  1.) 
But  to  take  successively  negative  steps  of  5  and  7 
units  respectively  amounts  to  taking  a  single  step  of 
12  negative  units.     (See  Fig.  1 ;  also  Ex.  2,  above.) 

20.  We  may  show  that  the  principles  applied 
above  may  be  extended  to  all  positive  and  nega- 
tive numbers,  by  representing  by  a  and  h  any 
arithmetic  numbers  whatsoever,  or  in  symbols 
as  below  : 

(i.)    +a  + +^  = +(^  + /v). 
(ii.)   -«  +  -/>  =  -(«  + ^). 

Proof  of  (i.): 

+«  +  +&  =  +(a  +  6). 

The  positive  units  represented  by  +6  taken  together 
with  the  positive  units  represented  by  +a  form  a 
combination  or  group  of  positive  units  represented  by 
+(^1  +  6). 

(ii.)    may  be  proved  by  similar  reasoning. 

21.  We  may  now  state  the  following 
Principle :    The  sum  of  two  algebraic  nvm- 

bers  of  like  sign  is  an  algebraic  number  of  the 
same  sign.  It  may  be  found  by  adding  arith- 
metically the  absolute  values  of  the  two  numbers 
entering  into  it. 

(b.)  The  numbers  to  be  added  may  be  of 
opposite  quality,  say 

One  Positive  and  the  Other  Negative. 

22.  Abstract  numhers  are  those  which  stand  alone.  They 
may  be  thought  of  as  "unnamed  numbers,"  that  is,  as  number 
names,  such  as  4,  5,  10,  etc.,  taken  by  themselves  without  reference 
to  any  particular  objects. 


£       ,, 

n2 

ni 

V       *' 

no 

/         o 

*9 

+  8 

T       ' 

*7 

^6 

/        < 

*5 

5        ^ 

*4 

+3 

0        ' 

^2 

P         o 

n 

^0 

-o'l 

-1   Y 

j^ 

-2    1 

E 

-3    T 

-4.    ? 

G 

-5    Y 

A 

-6    Y 

-7     7 

T 

-8    Y 

-9    Y 

I 

-10    Y 

V 

-11    T 

-12    ? 

E 

1 

Fig. 

1. 

POSITIVE  AND   NEGATIVE   NUMBERS  35 

23.  Concrete  numbers  are  those  formed  by  applying  the  num- 
ber names  to  particular  things. 

Concrete  numbers  are  "  named  numbers,"  as  5  oranges,  4  days,  etc. 

Positive  and  negative  numbers  are  "  named  numbers  "  and  behave  like 
concrete  numbers. 

As  things  unlike  in  kind  cannot  be  "added"  or  "united"  to  represent 
any  number  of  things  of  either  kind  alone,  so  positive  and  negative  num- 
bers, as  such,  are  to  be  looked  upon  as  being  entirely  separate  and  distinct, 
but  with  this  conditional  difference  always,  that  each  "offsets"  or  "de- 
stroys "  the  effect  produced  by  the  other  upon  any  particular  number  or 
quantity. 

24.  Hence,  positive  and  negative  numbers  equal  in  absolute 
value,  combined  by  addition,  may  be  neglected  in  a  series  of  addi- 
tions and  subtractions  since  together  they  produce  no  change  in  the 
total  result.  If  either  occurred  alone,  it  would  produce  a  change, 
but  both  together  act  in  such  a  way  as  to  "  oppose  "  each  other. 

E.  g.   Since  +10  +  "10  =  0,  +  +15  +  +10  +  "10  =  +15. 
Also,  since    -  +8  -  "8  =  0,       -f  +12  -  +8  -  "8  =  +12. 
and  in  general  +  +a  +  ~a  =  0,  also  —  +a  —  ~a  =  0. 

Ex.  3.    To  +9  add  "2. 

If  instead  of  saying  "  added  to,"  we  use  "  taken  together  with,"  in  the 
addition  of  algebraic  numbers  of  opposite  quality,  the  student  will  find  that 
the  idea  suggested  does  not  involve  the  difficulty  which  is  frequently  en- 
countered because  of  the  idea  of  an  increase  which  is  commonly  associated 
with  the  word  "addition." 

When  combined  together  into  one  whole,  the  two  negative  units  reduce 
the  number  of  positive  units  from  9  to  7. 

Hence  the  result  of  the  combination  is  7  positive  units,  and  we  may  write 

+9  +  -2  =  +(9  -  2)  =  +7. 

Ex.  4.  To  -7  add  +4. 

The  effect  of  4  positive  units  in  combination  with  7  negative  units  is  to 
reduce  the  number  of  effective  negative  units  by  4.  Hence  the  result  of 
the  "addition"  or  combination  is  3  negative  units.     We  may  write 

-7  +  +4  =  -(7  -  4)  =  -3. 

25.  It  will  be  noticed  that  in  the  examples  above  the  number  which  is 
greater  in  absolute  value  is  the  one  which  determines  the  quality  of  the 
result. 


36  FIRST  COURSE  IN  ALGEBRA 

26.  It  may  be  observed  tbat  the  reason  why  the  operation  which  appears 
to  be  subtraction  comes  under  the  head  of  addition  is  the  fact  that  we  are 
operating  with  numbers  of  opposite  quality,  that  is,  with  positive  and  nega- 
tive numbers.  The  "  apparent "  subtraction  is  not  strictly  addition  itself, 
but  belongs  rather  to  the  subsequent  reduction. 

27.  The  Commutative  L.aw  for  Addition,  that  is,  the  value 
of  a  sum  does  not  depend  upon  the  order  of  adding  its  parts^  may  be 
shown  to  hold  for  both  positive  and  negative  numbers. 

In  sjrmbok  +a  4-  "*"&  =  +  ^ft  +  "^a 

Also  +a  +  "ft  =  +  -6  +  ^a. 

E.  g.  +6  +  +.3  =  +3  +  +6.         +7  +  -5  =  -5  +  +7. 

-7  +  -4  =  -4  +  -7.         -2  +  +8  =  +8  +  "2. 

28.  From  the  consideration  of  the  preceding  examples  we  may 
state  the  following  General  Principles  : 

Principle  I. 

(i.)  The  sum  of  two  or  more  positive  numbers  is  a  positive  number 
whose  absolute  value  is  found  by  taking  the  sums  of  the  arithmetic 
values  of  the  positive  numbers  entering  into  it. 

In  symbols  "♦"«  +  "^6  =  "^(a  +  h). 

E.  g.  +5  +  +8  =  +(5  +  8)  =  +13. 

(ii.)  The  sum  of  two  or  more  negative  numbers  is  a  negative  num- 
ber whose  absolute  value  is  found  by  taking  the  sum  of  the  arithmetic 
values  of  the  negative  numbers  entering  into  it. 

In  symbols  -«  +  -&  =  -{a  +  &). 

E.  g.  -7  +  -3  =  -(7  +  3)  =  -10. 

29.  Principle  II.  The  algebraic  sum  of  several  positive  and 
negative  numbers  is  found  by  taking  the  arithmetic  difference  be- 
tween the  absolute  value  of  the  sum  total  of  the  positive  numbers  and 
the  absolute  valus  of  the  sum  total  of  the  negative  numbers^  and  it 
agrees  in  quality  with  the  greater  sum  total. 

In  symbols        "^a  +  ~6  =  +(«  —  6),  for  a>b. 
Or  =  ~(h  —  «),  for  a  <b. 


POSITIVE   AND   NEGATIVE   NUMBERS  37 

Ex.  5.    +5  +  +2  +  -6  +  -1  +  -3  +  +8  =  +(5  +  2  +  8)  +  -(6  +  1  +  3) 

=  +(15) +-(10) 
=  +5. 

Ex.  6.    +3  +  -6  +  -12  +  +4  +  -3  =  +(3  +  4)  +  -(6  +12  +  3) 

B+(7)+-(21) 
=  -14. 


30.  By  giving  concrete  meanings  to  the  positive  and  negative  numbera 
appearing  above,  for  example,  letting  the  positive  numbers  stand  for  items 
of  income  and  the  negative  numbers  for  items  of  expense,  it  will  appear 
that,  taken  together,  the  balance  will  be  $5  in  favor  of  the  income  in 
the  first  example,  while  in  the  second  example  there  remains  a  total  unpaid 
debt  of  $14. 


Exercise  IV.    2 


Simplify  the  following  expressions: 

1.    +2  +  +5. 

12. 

+5  +  -13. 

23. 

-18  +  +16. 

2.    +4  +  +3. 

13. 

-6  +  +14. 

24. 

+19  +  +19. 

3.    +6  +  +8. 

14. 

+7  +  -17. 

25. 

+16  4--10  +  -4. 

4.    +9  +  +6. 

15. 

+12  +  -12. 

26. 

+11  +  +18 +  -19. 

5.    -5  +  -7. 

16. 

+13  +  -14. 

27. 

+1  +  -3  +  +20. 

6.   -8  +  ~10. 

17. 

+15  +  +16. 

28. 

+7  +  -13  +  +6. 

7.    +7  +  -4. 

18. 

-17  +  "8. 

29. 

+12  +  -5  +  -7. 

8.    +10 +  -2. 

19. 

-14  +  -20. 

30. 

+13  +  -14  +  +15. 

9.    +11  +  "9. 

20. 

-4  +  -19. 

31. 

+3  +-6  +  +10. 

10.    +3  +  -11. 

21. 

-16  +  -3. 

32. 

+9  +  -17  +  -20. 

11.    +1  +  -6. 

22. 
II 

-19  +  +1. 
.   Subtraction 

31.    (i.)  Subtraction  of  Positive  Numbers. 

To  add  two  numbers  in  arithmetic  is  to  take  a  "positive"  step ; 
that  is,  a  step  "  upward  "  along  the  number  series  in  the  direction 
of  the  increasing  values  1,  2,  3,  etc. 

Hence,  to  subtract  a  positive  number  results  in  a  change  of  posi* 


38  FIRSl    COURSK  IN  ALGEBRA 

tion  along  the  number  series  in  a  "  downward  "  direction ;  that  is, 
in  the  direction  of  the  decreasing  values  8,  7,  6,  5,  etc. 

A  decrease  of  positive  values  occurs  also  whenever  we  take  a  negative 
step,  which  is  one  taken  in  the  direction  4,  3,  2,  1.  (See  Chapter  III, 
§13.) 

32.  "Taking  away,"  or  subtracting,  a  positive  step  amounts  to 
the  operation  of  adding  a  negative  step. 

E.  g.  Using  our  scale  of  extended  number,  we  may  understand  subtract- 
ing or  "  taking  away  "  +6  as  mcianing  counting  "  backward  "  or  "  downward  " 
in  the  direction  +10,  +9,  +8,  +7,  etc.,  for  a  distance  of  6  units,  beginning 
with  some  point  such  as  0,  until  we  finally  reach  -6.  By  adding  a  nega- 
tive step  of  6  units  we  shall  arrive  at  the  same  point.  Hence,  we  may 
regard  the  subtraction  of  6  positive  units  as  amounting  to  the  addition  of 
6  negative  units,  and  write 

-+6  =  +-6, 
and  in  general,  — +a  =  +~a. 

A  movement  "  downward  "  along  the  number  series  in  the  direc- 
tion ■•'lO,  "^9,  "•'8,  "^7,  etc.,  which  takes  place  among  the  positive 
numbers  alone,  that  is,  wholly  "  above "  zero,  corresponds  to  an 
arithmetic  subtraction. 

33.  (ii.)   Addition  of  Negative  Numbers, 

A  step  "downward,"  which  takes  place  "below"  zero,  that  is, 
among  the  negative  numbers  alone,  amounts  to  an  addition  of  neg- 
ative values. 

A  negative  step  is  taken  in  a  "  downward  "  direction,  that  is,  in 
the  direction  of  the  diminishing  values  +4,  +3,  '''2,  "^1. 

Hence,  removing  or  subtracting  a  "  downward  "  step  amounts  to  a 
change  in  position  in  an  "  upward  "  or  positive  direction ;  that  is, 
the  subtraction  of  a  negative  amounts  to  the  addition  of  a  positive 
step  or  number. 

That  is,  —-a  =  -f +«. 

E.  g.  --4  =  ++4. 


POSITIVE  AND   NEGATIVE   NUMBERS 


39 


34.   Correspondence  between  the  Addition  and  Subtraction 
OF  Positive  and  Negative  Numbers 


1  *^ 

(i)  Addition 
of  Posi- 
tives. 
Step 

1 

o    '1 

(ii)  Subtrac- 
tion of 
Positives. 
Step 

■^0 

(1    '^ 

>*' 

-0    ^ 

0 

-1 

-2 
-3 


(iv)  Subtrac- 
tion of 
Negatives. 
Step 


-1 

-2 
-3 


(iii)  Addition 
of  Neg- 
atives. 

Step 


Fio.  2. 


Reversing  a  "downward  "  step 
changes  it  into  an  "upward" 
step.  Hence,  subtractinj?  a 
negative  aiuoiiuts  to  add- 
iiij?  a  positive;  that  is  : 

—-(i  =  +  +rt.     (See  Fig.  2.) 


Also,  reversing  an  "  upward" 
step  changes  it  into  a  "down- 
ward "  step.  Hence,  subtract- 
iuj?  a  positive  ainoiints  to 
addinjiT    a    iiegrative,   or : 

-+a  =  +-a.     (See  Fig.  2.) 


35.  Hence,  for  positive  and  negative  numbers  we  have  the 
following 

Principle :  Every  operation  of  subtraction  of  positive  or  of  negative 
numbers  may  be  replaced  by  an  equivalent  addition  ;  that  is,  by  the 
addition  of  a  number  equal  in  absolute  value,  but  of  reversed  quality. 

Hence,  to  subtract  one  number  from  another,  reverse  the  sign  of 
quality  of  the  subtrahend  from  +  to  —  or  from  —  to  -f ,  and  then 
proceed  as  in  addition. 

In  symbols  :"  +a  -  +6  E  +a  +  ~h, 

36,  The  principles  for  the  removal  and  insertion  of  parentheses 
'i^ply  also  to  positive  and  negative  numbers. 


40  FIRST  COURSE  IN  ALGEBRA 

It  follows  that  the  Laws  of  Commutation  and  Association  for 
successive  additions  hold  also  for  successive  subtractions,  or  for  a 
chain  containing  both  additions  and  subtractions. 

Ex.  1.   From  +7  subtract  +2. 

We  may  express  the  operation  by  writing  +7  —  "^2. 

Replacing  the  operation  of  subtraction  by  that  of  addition,  reversing  at 
the  same  time  the  sign  of  quality  before  the  2,  we  have  +7  —  +2  =  +7  +  -2. 

When  combined  with  7  positive  units,  2  negative  units  reduce  the  number 
of  positive  units  by  2,  producing  as  a  result  of  the  combination  or  "  addi- 
tion," 5  positive  units. 

Hence,  the  expression  above  reduces  to  +(7  —  2)  =  +5. 

We  may  obtain  the  same  result  by  another  method  as  follows: 

Since  7  positive  units  amount  to  5  positive  units  increased  by  2  positive 
units,  we  may  write 

+7  _  +2  =  +5  +  +2  -  +2. 

Since  adding  and  subtracting  the  same  number  to  or  from  +5  produces  no 
change  in  the  final  result,  that  is,  since  4-  +2  +  -2  =  0,  we  may  write 

+5  +  +2  _  +2  as  +5. 

Ex.  2.   From  +3  subtract  +8. 

We  may  express  the  subtraction  by  writing  +3  —  +8.  Replacing  the 
operation  of  subtraction  by  that  of  addition,  reversing  at  the  same  time  the 
sign  of  quality  of  the  subtrahend,  we  have  +3  —  +8  =  +3  +  ~8. 

Since  in  combination  the  3  positive  units  reduce  the  number  of  negative 
units  from  8  to  5,  we  may  write  +3  +  "8  as  -(8  —  3)  =  ~5. 

We  may  also  employ  the  following  method : 

Since  subtracting  8  positive  units  in  one  operation  amounts  to  subtract- 
ing successively  3  positive  units  and  5  positive  units  in  two  separate  opera- 
tions, we  may  write  both  as 

+3  _  +8  =  +3  -  +3  -  +5. 
Observing  that         +3  —  +3  =  0, 
we  have  as  a  result  +3  —  +8  =  —  +5  =  +  -5. 

Ex.  3.    From  +6  subtract  -4. 

We  may  indicate  the  operation  by  writing  +6  —  "4.  Since  taking  away 
a  negative  amounts  to  adding  a  positive,  we  may  replace  the  indicated  sub- 
traction by  an  equivalent  addition,  and  write 

+6  -  -4  =  +6  +  +4  =  +(6  +  4)  =  +10. 


POSITIVE  AND   NEaATIVE  NUMBERS  41 

Ex.  4.    From  +5  subtract  -9. 

Observe  that,  since  the  subtraction  of  a  negative  amounts  to  the  addition 
of  a  positive,  we  may  from  the  indicated  subtraction  +5  —  "9,  obtain 

+5  -  -9  =  +5  +  +9  =  +(5  +  9)  =  +14. 

Ex.  5.    From  -12  subtract  -3. 

Writing  first  as  an  indicated  subtraction,  and  then  transforming  into  an 
equivalent  addition,  we  have  —12  —  — 3  =  ~12  +  +3. 

Since  in  combination  with  the  12  negative  units  the  3  positive  units 
diminish  the  number  to  9  negative  units,  we  may  write 

-12  +  +3  as  -(12  -  3)  =  "9. 

From  another  point  of  view,  12  negative  units  may  be  obtained  by 
combining  9  negative  units  with  3  negative  units. 

Hence  -12  —  "3  =  "9  +  -3  — -3;  and  since  successively  adding  and 
subtracting  3  negative  units  produce  no  final  change  in  the  value  of  the 
original  9  negative  units,  the  second  member  of  the  identity  reduces  to  "9. 

Ex.  6.   From  -4  subtract  -11. 

Replacing  the  indicated  subtraction  by  an  equivalent  addition,  we  have 
-4_-ll  =-4 +  +11. 

In  combination  with  the  11  positive  units  the  4  negative  units  produce  a 
decrease  in  number  to  7  positive  units.  Hence,  — 4  +  +11  may  be  written  as 
+(11  _4)  =  +7. 

From  another  point  of  view,  the  subtraction  of  11  negative  units  in  one 
operation  amounts  to  the  two  separate  operations  of  subtracting  4  and  7 
negative  units  successively.     Hence,  we  may  write  — 4  —  -1 1  =  -4  —  —4  —  —7. 

Since  the  subtraction  of  4  negative  units  from  4  negative  units  produces 
zero,  we  have  as  a  remainder  tlie  expression  — —7,  which  may  be  transformed 
into  the  equivalent  expression  +  +7. 

Hence  -4  —  -1 1  =  +  +7,  as  above. 

Ex.  7.   From  "1  subtract  +14. 

The  subtraction  of  14  positive  units  may  be  looked  upon  as  amounting  to 
the  addition  of  14  negative  units. 

Hence  -1  -  +14  =  -1  +  "14. 

In  combination  by  addition,  14  negative  units  and  1  negative  unit 
amount  to  15  negative  units. 

Hence  -1  +  -14  may  be  expressed  as  -(1  +  14)  =  -15. 

Ex.  8.    From  -16  subtract  +10. 

Observe  that  the  subtraction  of  10  positive  units  is  equivalent  to  the 
addition  of  10  negative  units.     We  may  write  -16  —  +10  =  "16  +  "10. 

We  have  an  expression  for  16  negative  units  increased  by  10  negative 


42  FIRST  COURSE   IN   ALGEBRA 

units,  resulting  in  26  negative  units.     Hence,  ~16  +  "10  may  be  written  as 
-(16+  10)  =-26. 

37.  luequality.  To  agree  with  the  ordinary  arithmetic  notions 
of  ineiiuality,  mathematicians  have  agreed  to  call  one  algebraic  num- 
ber, r/,  greater  or  less  than  another,  b,  according  as  the  reduced 
value  of  a  —  ^  is  positive  or  negative. 

From  the  definition  it  appears  that  any  positive  number  (repre- 
sented by  ^d)  must  be  considered  greater  than  any  negative  number 
(represented  by  ~b)  since 

.    -^a--b  =  -^a  +  -^b=  +(«  +  b) 

which  is  a  positive  number. 

From  the  same  point  of  view  0  is  to  be  regarded  as  being  greater 
than  any  negative  number,  ~/>,  since 

0  -  -6  =  0  -f  +^  =  +6 

which  is  a  positive  number. 

Hence,  corresponding  to  the  expression  "greater  than  0,"  there 
follows  directly,  by  application  of  the  Principle  of  No  Exception, 
also  the  idea  "less  than  0." 

38.  Again,  from  our  previous  definition,  one  negative  number, 
~c,  is  to  be  regarded  as  being  greater  than  another  negative  number, 
~dj  according  as  the  reduced  value  of 

~c  —  ~^  =  "c  +  '^d 
is  positive  or  negative. 

The  reduced  value  of  ~c  +  ^d  will  be  negative  if  c  be  numerically 
greater  than  ^,  and  positive  if  d  be  numerically  greater  than  c. 

Ex.  9.    Compare  "3  and  -4. 

— 3  —  -4  =  -3  +  +4  B  "^1,  a  positive  number. 
Hence  —3  is  greater  than  —4. 
Ex.  10.    Compare  —5  and  "l. 
Front  the  definition  we  have 

— 5  —  -1  B  — 5  +  +1  =  —4,  a  negative  number. 
Therefore  —5  is  less  than  ~1. 

39.  Using  the  symbol  qo,  "infinity,"  to  represent  any  number 
which  is  numerically  greater  than  any  assignable  number,  it  follows 


POSITIVE   AND  NEGATIVE   NUMBERS 


43 


from  the  reasoning  above  that  the  series  of  extended  number  may 
bo  regarded  as  being  arranged  in  order  of  increasing  magnitude, 
from  left  to  right,  as  follows: 

-^, ,  -3,  -2,  -1,  ±0,  +1,  +2,  +3, ,+co. 

Observe  that  0  may  be  regarded  as  belonging  to  both  the  positive 
and  negative  parts  of  the  series;  it  is  all  that  they  have  in  common. 
(See  Chapter  11.  §  23.) 

Exercise  IV.   3 

Perform  the  following  indicated  subtractions  : 


1.  From  +6  subtract  +4.  16. 

2.  From  +8  subtract  +1.  17. 

3.  From  +10  subtract  +7.  18. 

4.  From  +12  subtract  +5.  19. 

5.  From  "9  subtract  +3.  20. 

6.  From  "4  subtract  +2.  21. 

7.  From  "7  subtract  +8.  22. 

8.  From -11  subtract  "12.  23. 

9.  From  "5  subtract  "11.  24. 

10.  From  "2  subtract  +14.  25. 

11.  From  "3  subtract +19.  26. 

12.  From  -14  subtract  +9.  27. 

13.  From  "17  subtract  +15.  28. 

14.  From  +13  subtract  "20.  29. 

15.  From  -18  subtract  +16.  30. 


From  -1  subtract  +17.- 
From  -15  subtract  +18. 
From  +19  subtract  -19. 
From  -16  subtract  +16. 
From  -15  subtract  +14. 
From  -9  subtract  +11. 
From  -12  subtract  +13. 
From  +20  subtract -17. 
From  +18  subtract  +20. 
From  "6  subtract  +20. 
From  -7  subtract  +18. 
From  -8  subtract  +14. 
From  -9  subtract  +13. 
From  +10  subtract  -12. 
From  -11  subtract  +11. 


Perform  the  following  indicated  additions  and  subtractions  : 


31. 
32. 
33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 


+2  +  +3  -  +4. 
+1  -  +5  -  +15. 
+5  -  +9  +  +16. 
+5  +  -9  -  -16. 
+5  +  +9  -  +16. 
+3  —  +7  -  +9. 
+11  -+15  — -2. 
+17  —  -3  +  -19. 
+10  +  -19  --17. 
+20  -  -13  +  +6. 


41. 
42. 
43. 
44. 
45. 
46. 
47. 
48. 
49. 
50. 


+5  -  -5  +  -8. 
+13 --13 --19. 
+1  —+10  -  +19. 
+14  —  -8  +  +8. 
+16  +  -18- +9. 
+2  -  -4  +  +6  -  -8. 
+20  —  -18  +  -9  +  +6. 
+35  —  -15  +  -25. 
+40  —  -20  —  +30. 
+75  +  -50  -  +25.- 


44  FIRST   COURSE   IN   ALGEBRA 

Find  the  values  of  the  following  expressions  when  the  given 
values  are  substituted  for  the  letters  appearing  in  them. 
If  a  =  +l,b  =  -5,  c  =  -S,d  =  +4. 

51.  a  +  b  +  c  +  d.  56.  a  -  [b  -  (c  +  d)]. 

52.  (a  +  b)-(c  +  d).  57.  ^[(a  +  b)  -  c]  -  d. 

53.  a-(b  +  c)  +  d.  58.  (+6  -a)  -  (+6  -  b). 

54.  a   +  (b-c)-  d.  59.  (c  -  +3)  -  (d  -  -4). 

55.  a-[b+  (c-  d)].  60.  (b  -  +5)  -  (c  -  -3). 


POSITIVE  AND  NEGATIVE  NUMBERS  45 


CHAPTER  V 

MULTIPLICATION  AND  DIVISION  OF  ALGEBRAIC  NUMBERS 

I.   Multiplication 

1.  The  multiplication  of  abstract  numbers  is  first  defined  in 
arithmetic  as  the  taking  of  one  number  as  many  times  as  there 
are  units  in  another. 

E.  g.  To  multiply  4  by  3,  we  take  as  many  4's  as  there  are  units  in  3. 
We  have  4x3  =  4  +  4  +  4=  12. 

2.  In  algebra,  as  in  arithmetic,  we  call  the  number  multiplied 
the  multiplicand,  the  number  which  multiplies  it  the  multiplier, 
and  the  result  of  the  operation  the  product. 

Whenever  the  first  of  two  numbers  such  as  2  x  3  is  regarded  as  the 
multiplier,  it  is  customary  to  read  the  product  as  "  2  times  3/'  while  if  3  is 
regarded  as  the  multiplier  and  2  as  the  multiplicand,  we  may  say  "2  multi- 
plied by  3." 

3.  Our  first  idea  of  multiplication  is  that  it  is  an  abbreviated 
addition. 

From  this  point  of  view  the  multiplier  must,  in  the  original  sense 
of  the  word,  be  the  result  of  counting ;  that  is,  it  must  be  a  positive 
whole  number. 

The  multiplicand  may  be  any  number  previously  defined,  that  is, 
it  may  be  abstract  or  concrete,  positive  or  negative,  or  even  zero, 
but  tlie  multiplier  must  be  an  abstract  number. 

In  a  product  consisting  of  two  abstract  numbers,  the  one  at  the 
right  is  usually  regarded  as  the  multiplier.  However,  since  we 
speak  commonly  of  2  books,  3  apples,  etc.,  mentioning  the  multiplier 
first,  mathematicians  find  it  convenient  to  arrange  a  given  product 
containing  numbers  and  letters  so  that  the  numerical  parts  shall 
occur  in  the  first  place. 


46  FIRST  COURSE   IN   ALGEBRA 

Repeating  a  quantity  does  not  alter  its  nature ;  hence  it  follows 
that  a  product  must  be  of  the  same  nature  as  the  multi- 
plicand. Hence  the  product  will  be  an  abstract  or  a  concrete 
number,  according  as  the  multiplicand  is  an  abstract  or  a  concrete 
number. 

Our  first  conception  of  a  product  can  have  no  meaning  when 
fractional,  negative,  or  zero  multipliers  are  considered,  for  from  the 
original  definition  we  can  multiply  by  positive  whole  numbers  only. 
Therefore  we  apply  the  Principle  of  No  Exception,  and  declare  that 
since  negative  numbers  have  the  forms  of  dififerences,  multiplica- 
tions with  them  may  be  performed  exactly  as  with  real  or  actual 
differences. 

4.  To  allow  of  carrj-ing  out  the  operation  of  multiplication  when 
the  multiplier  is  a  fraction  or  a  negative  number,  it  becomes  neces- 
sary to  give  an  extended  definition  of  multiplication : 

To  multiply  one  number  by  a  second  number  is  to  d^}  to  the  first 
number  that  which  must  be  done  to  the  positive  unit  ^\  to  obtain  the 


This  does  not  contradict  our  first  definition  as  given  in  arithmetic, 
but  includes  it  in  the  more  general  statement. 

E.g.   3  =  -1-1 +  1-1-1. 

Hence,  doing  the  same  thing  to  4  that  was  done  to  4-1  to  obtain  3,  we 
may  write  as  the  product  of  4  and  3 

4x3  =  +  4  +  4  +  4=12.     (See§l.) 

5.  This  general  definition  may  be  regarded  as  including,'  the  multiplica- 
tion of  fmctions.  Thus,  if  we  wish  to  multiply  |  by  f,  we  must  do  to  | 
that  which  was  done  to  unity  to  obtain  ^;  that  is,  we  must  divide  |  into 
seven  equal  parts  and  take  three  of  these  equal  parts  as  summands.     Each 

2 
of  the  equal  parts  of  |  will  be  r-71^,  and  by  taking  three  of  these  parts  as 

0*7 

summands  we  shall  have 

2  2         2-36 


®(?) 


7"^5-7'5-7      5-7      35 


6.  Just  as,  in  arithmetic,  numbers  result  from  repetitions  of  unity, 
so  positive  and  negative  numbers  may  be  regarded  as  resulting  from 
repetitions  of^^^  unit  o/positive  numbers  '^1,  and  the  unit  of  negative 
numbers  ~1.  ^ 


POSITIVE   AND  NEGATIVE  NUMBERS  47 

We  may  take  as  quality  units  the  numbers  +1  and  ~1. 

7.  In  the  multiplication  of  positive  and  negative  numbers  we  may 

have 

I.    Tlie  Multiplier  Positive. 

Ex.  1.    Multiply  +5  by  +3. 

By  the  extended  definition  of  multiplication  the  product  may  be  obtained 
by  operating  upon  the  multiplicand  +5  in  exactly  the  same  way  that  we  must 
operate  upon  the  unit  of  positive  numbers  +1  to  obtain  the  multiplier  +3. 

From  the  detinition  of  a  positive  number,  +3  =  +1x3. 

That  is,  we  obtain  the  multiplier  +3  by  multi[)lying  the  unit  of  positive 
numbers  +1  by  3,  retaining  its  quality  as  a  positive  number. 

Hence  to  multii)ly  +5  by  +3  we  retain  the  quality  of  the  multiplicand  +5 
and  multiply  its  absolute  value  by  3. 

That  is,  +5  X  +3  =  +(5  x  3)  =  +15. 

Ex.  2.    Multiply  —2  by  +4. 

Rt?asoning  as  before,  the  jjroduct  may  be  obtained  by  multiplying  the 
absolute  value  of  the  multiplicand  by  4,  retaining  its  ([uality  as  a  negative 
number. 

Hence,  "2  x  +4  =  -(2  x  4)  =  "8. 

8.  In  the  multiplication  of  positive  and  negative  numbers  we 
may  have 

II.    The  Multiplier  Negative. 

Ex.  3.    Multiply  +6  by  "3. 

By  the  extended  detinition  of  multiplication,  the  product  may  be  obtained 
by  performing  upon  the  multiplicand  +6  exactly  those  operations  which 
must  be  performed  ujjou  the  unit  of  positive  numbers  +1  to  produce  the 
multi[)lier  —3. 

From  the  definition  of  a  negative  nund)er  we  have 

~3  =  4-  ~1  +  ~1  +  ~1,  ill  terms  of  negative  units, 
or  ~3  =  —  +1  —  +1  —  +1,  in  terms  of  positive  units. 

That  is,  -3  may  be  obtained  from  the  unit  of  positive  numbers  by  revers- 
"J  the  quality  of  the  positive  unit  and  multiplying  its  absolute  value  by  3. 

It  follows  that,  to  obtain  the  desired  product  of  +6  and  -3,  we  may  reverse 
the  quality  of  the  positive  unit  and  multiply  its  absolute  value  by  3. 

Hence,  +6  x  "3  =  "(6  x  3)  =  "18. 

ICx.  4.    Multiply  "7  by  -5. 

Reasoning  as  above,  we  may  reverse  the  sign  of  quality  of  the  multipli- 
cand -7  and  multiply  its  absolute  value  by  5. 

Hence  -7  x  "5  =  +(7  x  5)  =  +35. 


48  FIRST  COURSE  IN   ALGEBRA 

9.  It  should  be  observed  that,  in  each  of  the  examples  above,  the 
sign  of  quality  of  the  product  is  positive  or  negative  according  as 
the  signs  of  quality  of  the  multiplicand  and  multiplier  are  like  or 
unlike. 

10.  "We  will  now  show  that  this  Law  of  Signs  holds  for  all  positive 
and  negative  numbers. 

Representing  by  a  and  h  any  two  arithmetic  numbers,  that  is,  integral 
values, 

(i.)   +a  X  +ft  =  +{ah),  (iii.)  -a  x  +6  =  -(nb), 

(ii.)  -ax-h  =  ^{ah)y  (iv.)  +a  x  "6  = -(a6). 

Proofs  of  (i.)  and  (iii.)  : 

To  obtain  +6  from  the  unit  of  positive  numbers  +1,  we  retain  the  quality 
of  the  unit  and  multiply  its  absolute  value  by  6 ;  it  follows  from  the  extended 
definition  of  multiplication  that  to  multiply  any  number  by  +6  we  retain 
the  quality  of  the  number  and  multiply  its  absolute  value  by  h. 

Hence,  (i.)   +a  x +h  =  +(a6). 
(iii.)   -a  x  "•'6  =  ~(ab). 
Proofs  of  (ii.)  and  (iv.)  : 

To  obtain  -h  from  the  unit  of  positive  numbers  +1  we  may  change  the 
sign^  of  quality  of  the  unit  and  multiply  the  result  by  b ;  accordingly,  to 
multiply  any  number  by  "6,  we  may  reverse  the  quality  of  the  number  and 
multiply  the  result  by  b. 

Hence  (ii.)  -a  x  ~b  =  +(a&). 
(iv.)    +a  X  ~b  =  -(ab). 

The  Law  of  Signs  for  the  multiplication  of  two  numbers  or  quan- 
tities may  be  stated  as  follows  : 

TTie  ])7vduct  of  two  numbers  having  like  quality  signs  is  positive, 
and  the  product  of  two  numbers  having  unlike  quality  signs  is 
negative. 

11.  By  examining  identities  (i.)  to  (iv.),  §  10  above,  it  may  be 
seen  that  the  signs  of  quality  of  both  multiplicand  and  multiplier 
may  be  reversed  without  altering  the  sign  of  quality  and  numerical 
value  of  the  product. 

E.  g.  Reversing  the  signs  of  multiplicand  and  multiplier  in  (i.),  we  obtain 
the  multiplicand  and  multiplier  in  (ii.),  but  the  quality  of  the  product 
remains  unaltered. 


POSITIVE  AND   NEGATIVE  NUMBERS  49 

Reversing  the  signs  of  quality  of  multiplicand  and  multiplier  in  either 
(iii.)  or  (iv.)  has  the  effect  of  an  interchange  of  signs,  giving  as  a  result 
the  left  member  of  either  (iv.)  or  (iii.),  the  quality  and  numerical  value  of 
the  product  remaining  unaltered  by  the  change. 

12.  Hence  we  have  the  Commutative  Law  for  Signs  of 
Quality  in  Multiplication,  that  is, 

~a  X  ^b  H  ^a  X  ~b  =  -(ah). 

It  follows  that,  in  establishing  the  Commutative  Law  for  the 
multiplication  of  positive  and  negative  numbers,  we  may  establish 
it  for  the  last  form  ~{ab)  only,  using  the  absolute  values  of  a  and  b. 


Exercise  V.   1 

Simplify  the  following  : 

1.  +3  X  +2. 

10.  +12  X  -8. 

19. 

-10  X  -19. 

2.  +5  X  +4. 

11.  -11  X  -10. 

20. 

+18  X  -10. 

3.  +7  X  +3. 

12.  -14  X  "6. 

21. 

+15  X  +1. 

4  +2  X  +9. 

13.  -4  X  -15. 

22. 

-1  X  -14. 

5.  -^9  X  +5. 

14.  "2  X  -17. 

23. 

+19  X  -4. 

6.  +4  X  -7. 

15.  -15  X  -3. 

24. 

+9  X  +12. 

7.  +6  X  -11. 

16.  -16  X  -5. 

25. 

-7  X  +9. 

8.  +8  X  -12. 

17.  +6  X  -20. 

26. 

-6  X  -8. 

9.  +10  X  -13. 

18.  -3  X  +18. 

27. 

-20  X  -15. 

28.  (+2  X  +5)  +  (-5  X  +6).  34.   ("8  X  "7)  -  (-7  X  +8). 

29.  (+11  X  -11)  +  (+13  X  -2).  35.  (+3  X  "4)  -  (-5  X  +6). 

30.  (+5  X  -16)  ~  (+4  X  +3).  36.  (+7  X  +10)  -  (+8  X  +11). 

31.  (+1  X  -1)  -  (-20  X  +20).  37.   (+10  X  -14)  +  (+15  X  "15). 

32.  (-1  X  +3)  -  (+12  X  +12).  38.   (+13  X  +20)  -  (+20  X  "14). 

33.  (+16  X -16) -(-16  X  +16).  39.  (+10  X  -18)  +  (+18  X  +13). 

lia  =  +2,  6  =  -3,  x  =  ~b,  y  —  +4,  find  the  values  of  the  follow- 
ing expressions  : 

40.  ah  +  (cy.  45.   ah  +  y. 

41.  ax  +  by.  46.   x  —  ah. 

42.  ay  —  hx.  47.  y  —  hx. 

43.  ah  +  a.  48.   ah  +  hx  +  xy, 

44.  hx  +  X.  49.   ax  •\-  ah  ■\-  by. 


L 


50  FIRST  COURSE   IN  ALGEBRA 

Find  the  value  of  (a  +  b)  X  c,  when 

50.  a  =  +2,  6  =  +6,  c  =  +3.      52.   a  =  H,b  =  -5,c  =  "7. 

51.  a  =  -3,  6  =  +1,  c  =  +4. 
Find  the  value  of  (a  —  b)  X  c  when 

53.  a  =  -1,  6  =  +4,  c  =  "8.      55.   a  =  +10,  b  =  "10,  c  =  +10. 

54.  a  =  +6,  ^  =  +9,  c  =  "1. 

Find  the  value  of  (a  +  b)  X  (c  -\-  d)  when 

56.  a  =  -1,  6  =  "2,  c  =  +3,  </  =  +4. 

57.  a  =  +5,  ^^  =  -2,  c  =  +12,  c?  =  "9. 
Find  the  value  of  (a  —  b)  X  (c  —  d)  when 

58.  a  =  +S,b  =  +5,  c  =  "2,  d  =  "4. 

59.  a  =  +2,  6  =  +7,  c  =  "3,  c?  =  +11. 

13.  "Whenever  both  the  multiplicand  and  multiplier  are  abstract 
numbers,  two  fundamental  laws  hold  also  in  multiplication.  These 
are  the  Law  of  Commutation,  affecting  arrangement,  and  the  Law  of 
Association,  affecting  grouping.  These  laws  are  similar  to  the  laws 
in  addition  having  the  same  names.     Thus  : 

(i.)   a  times  b  =  b  times  a.     Law  of  Commutation. 

(ii.)   a  times  (b  times  c)  =  (a  times  b)  times  c.     Law  of  Association. 

14.  The  Commutative  Law  for  Multiplication 

In  a  product  of  two  abstract  numbers,  either  number  may  be  taken 
as  the  multiplier  without  affecting  the  value  of  the  result. 

Thus,  in  s)rmbols,  a  xh^hx  a. 

E.  g.  5x3  =  3x5. 

We  may  show  that  the  law  holds  for  the  product  of  two  particular  num- 
bers, say  5  and  3,  by  representing  the  number  5  by  five  points  in  a  horizontal 
row,  and  constructing  three  rows,  as  follows : 


Counting  the  entire  set  of  dots,  we  may  regard  it  as  consisting  of  three 
groups  of  5  dots  each,  written  5  x  3,  or  again  as  five  groups  of  3  dots  each, 
written  3x5.     Hence,  we  may  assert  that  5x3  =  3x5. 

If,  instead  of  reasoning  with  particular  numbers,  we  arrange  h  horizontal 


POSITIVE  AND  NEGATIVE   NUMBERS  51 

rows  of  a  dots  each,  we  shall  arrive  by  similar  reasoning  at  the  result,  that 
for  all  positive  integral  values  of  a  and  b 

a  X  b  =  b  y  a. 

This  is  called  the  Law  of  Commutation  for  Multiplication. 

It  may  be  shown  that  the  principle  applies  whenever  there  are 
three  or  more  factors. 

15.  Continued  Product. 

The  product  of  three  or  more  numbers,  a,  h,  c,  and  d^  written 
a  X  b  X  c  X  dy  ox  abcd^  is  defined  to  be  the  number  obtained  by- 
multiplying  a  by  6,  this  result  by  c,  and  finally  the  last  result  by  d. 

If  we  represent  any  three  arithmetic  whole  numbers  by  «,  &,  and  c,  we 
mav  indicate  the  continued  product  of  three  positive  numbers  by  writing 
(+a)(+6)(+c). 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  foj  the  first  time.) 
By    the    definition    of    a    continued    product    we    are    to    understand 
(+a)(+6)(+c)  as  meaning  that  we  are  to  first  multiply  +a  by  +6,  and  this 
product  by  +c. 

(+a)(+6)(+6')  =  [(+a)(+i)](+c).     By  definition  of  multiplication. 
But  (+«)(+&)  = +(a6). 

Hence     (+a)(+/>)(+c)  =  [+(a6)](+c). 

Regarding  +(«6)  as  a  single  number,  and  multiplying  by  +c, 

=  +(ak). 
By  assuming  some,  all,  or  none  of  the  three  factors  of  the  product  to  be 
positive  numbers,  we  may  extend  the  principle  to  include  such  combinations 
of  positive  and  negative  numbers  as  the  following : 

(+a)(+6)(+c)  =  +(a&)(+c)  =  +{ahc). 
{-a){+b){+c)  =-{ab){+c)  =-{abc), 
{-a)(-b){+c)  =  nfib){+c)  =  ^((ibc). 
{-€i)(-b){-c)  =  +{ab){-c}  =  -{abc), 
and  so  on  for  more  factors. 

16.  The  essential  thing  to  be  observed  in  the  identities  above  is 
that  a  continued  product  is  positive  if  it  contains  no  negative  factors 
or  if  it  contains  an  even  number  of  negative  factors,  and  it  is  nega- 
tive if  it  contains  an  odd  number  of  negative  factors. 

E.g.  1.  (+3) (+2) (+4) (+5)    =+120. 

^2.  (-4)(-6)(+l)(+7)   =+168. 

3.  (-l)(-3)(-8)(-10)=+240. 

4.  (-2)(+3)(+5)(+0)    =-180. 

5.  (-5)  (-2)  (-4)  (+9)    B-360. 


52  FIRST  COURSE   IN  ALGEBRA 

17.  The  Associative  Law  for  Multiplication 

The  value  of  a  product  remains  unaltered  ij\  in  the  process  of 
multiplying  several  numbers^  two  successive  factors  are  associated  or 
grouped  together  to  form  a  single  product. 

E.  g.    2  X  3  X  4  X  5  =  2  X  3  X  (4  X  5)  =  (2  X  3)  X  20  =  6  X  20  =  120. 
(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

For  any  three  arithmetic  whole  numbers,  a,  6,  c, 

ahc  =  a(bc). 
We  have  ahc  =  {ah)c.      By  definition  of  a  product. 

Considering  the  product  (ab)  as  one  number,  by  the  Commutative  Law 
for  two  factors  we  have : 

=  c(ab)f 

=  (ca)b,  by  definition  of  a  product. 
=  b(ca),  by  Commutative  Law. 
=  (bc)af  by  definition  of  a  product. 
=  a(bc),  by  Commutative  Law. 
Or,  ahc  =  {ah)c  =  a(bc). 

18.  By  repeated  applications  of  the  Commutative  and  Associative 
Laws  for  multiplication,  it  may  be  shown  that  both  laws  hold  for 
three  or  more  factors.     That  is  : 

abc  =  acb  =  hac  =  hca  =  cab  =  cba. 
Also,  abed  =  a  (bed)  =  a  (be)  d  =  b  (acd)  =  etc., 

and  so  on  for  any  number  of  factors. 

19.  From  the  above,  it  appears  that  we  may  arrange  the  factors  of 
a  product  in  any  order,  and  group  them  together  in  any  convenient 
way,  ivithout  altering  the  value  of  the  result. 

20.  Both  the  multiplicand  and  multiplier  receive  the  name  of 
factor,  since  they  may  be  interchanged  without  altering  the  value 
of  the  product. 

E.  g.   a  and  h  are  factors  of  the  product  a  X  6. 

Similarly,  each  number  of  a  continued  product,  ahcdef ,  is 

called  a  factor  of  that  product. 

E.  g.  5,  a,  6,  and  c  are  all  factors  of  the  continued  product  5  abc. 

21.  The  Distributive  Law  for  Multiplication 

Up  to  this  point  we  have  considered  products  in  which  both  mul- 
tiplicand and  multiplier  consisted  of  single  numbers.     In  case  eithei 


POSITIVE   AND   NEGATIVE  NUMBERS  63 

or  both  are  sums  or  differences,  we  are  led  to  consider  the  third 
Fundamental  Law  of  Algebra,  namely,  the  Distributive  Law. 

In  particular,  we  will  show  that  the  product  of  3  multiplied  by  the  sum 
of  4  and  5  is  the  same  as  the  product  of  3  multiplied  by  4,  increased  by  the 
product  of  3  multiplied  by  5. 

Let  a  series  of  dots  be  arranged  as  below,  forming  a  set  of  three  rows, 
each  containing  9  dots. 


The  dots  may  be  counted  in  either  of  two  ways :  first,  as  a  single  group 
consisting  of  three  rows  containing  9  dots  each,  that  is,  27  dots  in  all ; 
second,  as  consisting  of  one  group  of  three  rows  containing  4  dots  each, 
and  a  second  group  consisting  of  three  rows  containing  5  dots  each,  —  the 
two  groups  being  separated  as  shown. 

Hence  we  may  write 

3(4  +  5)  =  (3  X  4)  +  (3  X  5)  =  27. 

22.  The  process  may  be  applied  to  any  three  whole  numbers,  a, 
b,  c,  and  we  may  assert  as  a  general  principle  that 

The  product  of  an  algebraic  sum  multiplied  by  a  single  number  may 
be  obtained  by  multiplying  each  term  of  the  sum  by  the  given  number^ 
and  finding  the  algebraic  sum  of  the  results  obtained. 

Or  a{h  +  c)  =  ab  +  ac. 

This  is  called  the  Distributive  Law  for  Multiplication,  and  it  may 
be  shown  to  hold  when  the  multiplier  consists  of  any  number  of 
terms,  which  may  be  positive  or  negative,  integral  or  fractional. 

23.  Zero  as  a  Factor. 

It  follows  directly  from  the  Fundamental  Laws  that  a  product  is 
zei'o  if  one  of  its  factors  is  zero. 

That  is  a  •  o  =  o,     (Multiplier  0). 

O  •  a  =  O.     (Multiplicand  0). 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 


We  may  write 

n  - 

-  n  =  0,     as  defining  zero. 

Accordingly, 

a 

•  0  =  a(n-n)     (Multiplier  0). 

=  an  —  an    By  Distributive  Law  for  Mul- 
tiplication. 
=  0.     By  definition  of  0. 

54  FIRST  COURSE  IN  ALGEBRA 

Also,  0  '  a  =  (n  —  n)a     (Multiplicand  0.) 

=  na  —  na 
=  0. 

24.  Since  the  proof  of  an  identity  establishes  at  the  same  time 
the  truth  of  its  converse,  it  follows  that,  if  a  product  is  zero,  at  least 
one  of  its  factors  must  be  zero, 

that  is,  if  a  •  b  =  0, 

then  either  a  is  0,  or  h  is  0,  or  both  a  and  h  are  0. 

25.  The  product  obtained  by  using  the  same  factor  repeatedly 
is  called  a  power  of  that  factor. 

E.  g.  3  X  3  is  called  the  second  power  of  3,  or  3  raised  to  the  second 
power,  since  3  occurs  twice  as  a  factor. 

Also  2x2x2x2  is  called  the  fourth  power  of  2 ;  etc. 

26.  The  number  of  times  a  factor  appears  in  a  product  may  be 
indicated  by  writing  a  small  number  called  the  exponent  or  the 
index  of  the  power  at  the  right  of  and  immediately  above  the 
factor. 

E.  g.    We  may  write  5^  instead  of  5  x  5  ;  4^  instead  of  4  x  4  x  4  ;  etc. 

27.  The  number  which  is  used  repeatedly  as  a  factor  to  obtain  a 
power  is  called  the  base  of  the  power. 

28.  The  definition  of  an  exponent  as  given  is  that  it  indicates  the 
number  of  times  a  factor  appears  in  a  product.  This  definition 
requires  that  the  exponent  should  be  a  positive  whole  number. 

In  a  later  chapter  this  notion  of  an  exponent  will  be  somewhat 
extended. 

29.  In  arithmetic  a  number  is  defined  as  being  even  or  odd 
according  as  it  is  or  is  not  divisible  by  2. 

E.  g.  2,  4,  6,  10,  16,  etc.,  are  even  numbers. 

3,  6,  7,  11,  17,  etc.,  are  odd  numbers. 

30.  A  power  is  defined  as  being  even  or  odd  according  as  its 
exponent  is  even  or  odd. 

E-  g-  (+4)2,  (+6)^,  (-3)«,  (-7)8,  etc.,  are  even  powers, 

while  (+2)3,  (+3)5,  (-1)7,  (-2)9,  etc.,  are  odd  powers. 


POSITIVE  AND   NEGATIVE   NUMBERS  55 

31.  An  odd  power  of  a  negative  base  conta,ins  an  odd  number  of 
negative  factors,  and  accordingly,  by  the  rule  of  signs  for  continued 
products,  it  is  of  negative  quality. 

E.  g.    The  power  (~2)^  is  of  negative  quality. 
For,  r2)3=(-2)(-2)(-2)  =  -8. 

32.  Whenever  we  speak  of  a  positive  integral  power  we  have  ref- 
erence to  the  exponent  rather  than  to  the  value  of  the  base,  which 
may  itself  be  fractional  or  negative. 

E.  g.  Tlie  following  are  positive  integral  powers  of  fractional  bases  and 
of  negative  bases : 

©••  (I)".  «••  (-9' 

33.  In  operating  with  powers  we  are  governed  by  the  following 
Principles  : 

(i.)      All  powers  of  positive  bases  are  positive. 
(ii.)    Even  powers  of  negative  bases  are  positive. 
(iii.)    Odd  poivers  of  negative  bases  are  negative. 

34.  Since  a  product  is  zero  if  one  or  more  of  its  factors  is  zero,  it 
follows  that  any  positive  integral  power  of  zero  is  zero  ;  that  is, 

O"  =  O. 

35.  In  order  to  indicate  clearly  and  exactly  what  number  is  to  be 
considered  as  the  base,  it  is  often  necessary  to  enclose  the  number 
within  parentheses,  as  in  the  following  illustrations  : 

(i.)   Tlie  base  a  negrative  number. 

(-8r=(-8)(-3)  =  -^(3x:-0  =  -'9. 

Observe  that  ~a^  is  not  the  same  as  (CaY. 

~a^  is  read  "  negative  a  square ; "  (~a)^  is  read  "  the  square  of 
negative  a." 

We  have  ~a^  =  ~{a  X  a)  =  "a^, 

while  (-ay  =  (-a)(-a)  =-^a^ 

Whenever  the  symbol  before  the  number  or  base  is  regarded  as 
one  of  operation,  as  for  example,  — 3^  we  may  write 
-  3»  =  -  (3  X  3  X  3)  =  -27. 

(ii.)  Tbe  base  not  a  single  number,  but  either  a  product 
or  a  quotient. 


66  FIRST  COURSE   IN   ALGEBRA 

Ex.  1.  (2  X  5)2  =  (2  X  5)(2  X  5)  =  (10)(10)  =  100. 

The  example  above  should  be  distinguished  from  the  following: 
Ex.  2.  2  X  52  =  2(5  X  5)  =  2(25)  =  50. 

\4/    ~  4  ^  4  ~  16  * 


Ex.  3. 


If  the  numerator  alone  or  the  denominator  alone  is  to  be  raised  to 
power,  we  may  write 


32  _ 

9 

4 

■4* 

3  _ 

3 

42- 

'  16 

Similarly, 

(iii.)   The  base  a  sum  or  a  difference. 

Ex.  4.  (3  +  5)2  =  (3  +  5)(3  +  5)  =  (8)(8)  =  64. 

This  should  be  distinguished  from  the  following: 

3  +  52  =  3  +  (5  X  5)  =  3  +  25  rr  28, 
also  from  3^  +  5^  =  (3  x  3)  +  (5  x  5)  =  9  +  25  =  34. 

Ex.  5.  (5  -  3)2  =  (2)2  =  4. 

This  should  be  distinguished  from  the  following: 
52  -  32  =  25  -  9  =  16. 

(iv.)    The  base  a  power. 

Ex.  6.    (30*  =  (32)(32)(32)  =  (3  X  3)(3  x  3)(3  x  3)  =  9  x  9  x  9  =  729. 

The  use  of  exponents  above  should  be  distinguished  from  .32  ,  which  may 
be  taken  to  mean  either  (32)*  (read  "the  cube  of  the  second  power  of  3") 
or  3^2*^  (read  "  3  raised  to  the  power  two  cubed  "). 

That  is,  32*  =  (32)*  =  (32)  (32)  (32)  =  9  •  9  •  9  =  729, 

Or  32*  =  3^28)  ^  32x2x2  =  38  =  6561. 

Exercise  V.     2 
Find  the  values  of  the  following  indicated  powers  : 

Arithmetic  Numbers 

1.  2\  5.  2^  9.  1\  13.  5*. 

2.  2*.  6.  3*.  10.  6*.  14.  12*. 

3.  3^.  7.  4*.  11.  2«.  15.  9*. 

4.  51  8.  8^.  12.  3^  16.  10*. 


POSITIVE   AND   NEGATIVE  NUMBERS  57 

Positive  and  Negative  Numbers 

17.  (+2)«.  22.  (-3)1  27.  {nf.  32.  (+4)*. 

18.  (+4)2.  23.  (-6)2.  28.  (-2)^  33.  (-15)2. 

19.  (+5)^  24.  (-11)2.  29.  (-3)*.  34.  ("20)^ 

20.  (+8)2.  25.  (-9)«.  30.  (-4)».  35.  (+20)'. 

21.  (+10)^  26.  (-12)2.  31.  (-3)^  36.  (-13)2. 

Using  the  letters  «,  ^,  c,  ^,  y,  2;,  etc.,  to  represent  positive  whole 
numbers,  find  expressions  for  the  following: 

37.  (+2^0'-  40.    (-2.r)«.  43.    (+5aft)2.  46.    (+2«)2 

38.  (+3^)2.  41.    (-3^)2.  44.    (-4.r^)^  47.    (+32)2. 

39.  (+4c)^  42.    (-4;^)^  45.    ("3  6c)*.  48.    (-22)1 

Find  the  values  of  the  following  expressions  : 

49.  +2«  +  +32.         52.    +62  -  +32.         55.    +2^  -  -2l 

50.  +32  4-  +42.         53.   -5«  +  -^3^         56.    +12  +  +2^  +  +32  +  +42. 

51.  +52  -  +42.         54.   -102  ^  +92       57     +12  +  +32  ^  +52  ^  +72^ 
58.    +22 ---32 ++42 --52.  59.    (+42 -+52)2 -(-62 --72)2. 

II.  Division 

36.  The  terms  dividend,  divisor,  quotient,  remainder  are 

used  relatively  in  the  same  way  in  algebra  as  in  arithmetic. 

Division  as  an  operation  is  the  inverse  of  multiplication. 

To  divide  one  number  (dividend)  by  another  (divisor),  is  to  find 
another  number  (quotient),  which  when  multiplied  by  the  divisor 
produces  the  first  (dividend). 

E.  g.  To  divide  12  by  4  is  to  find  the  quotient  3.  Multiplying  the 
quotient  3  by  the  divisor  4  produces  the  original  dividend  12. 

37.  By  the  mutual  relation  of  multiplication  and  division  the 
quotient  has  the  fundamental  property  that,  when  multiplied  by 
the  divisor,  the  product  is  the  dividend. 

That  is,  Quotient  x  JDivisoi'  =  Dividend, 

If  we  represent  dividend,  divisor,  and  quotient  by  D,  d  and  Q  respectively, 
we  may  indicate  the  quotient  by  writing  -r ,  and  our  definition  of  division 
as  a  process  may  be  symbolized  by 

^xd  =  n. 

d 


58  FIRST  COURSE   IN   ALGEBRA 

38.  Division,  like  subtraction,  cannot  always  be  performed,  but 
it  may  always  be  indicated.  It  is  only  in  exceptional  cases  that 
there  can  be  obtained  an  integral  quotient  with  no  remainder.  In 
this  case  the  dividend  is  said  to  be  exactly  divisible  by  the 
divisor. 

39.  The  fractional  notation  for  a  quotient,  namely,  y*  and  the 

solidus  notation  «/6,  are  commonly  used  for  division.  Primarily, 
either  means  that  we  are  to  take  the  ^th  part  of  unity  a  times  as  a 
summand.  Hence,  h  times  a  of  the  h\h  parts  of  unity  is  equivalent 
to  a  times  unity ;  or  in  symbols, 

V  X  ft  =  o. 
o 

Also,  by  the  definition  of  division  we  have 

(a  -f-  ft)  X  6  =  a. 

Hence  y  has  the  same  meaning  as  a  -i-  ^  when  a  and  h  are  whole 

numbers. 

40.  When  division  can  be  performed  at  all,  it  can  lead  to  but  a 
single  result;  hence  it  is  called  a  determinate  process. 

Dimsion  by  0  is  not  an  admissible  operation.  > 

41.  Since  multiplication  and  division  are  mutually  inverse  opera- 
tions, it  follows  that  if  any  number  be  successively  multiplied  by, 
and  then  divided  by  the  same  number,  or  be  first  divided  by  and 
then  multiplied  by  the  same  number,  the  resulting  value  will  be 
the  same  as  though  no  operation  had  been  performed.     Or,  stated 

in  S3rmbols,  (a  -f-  &)  x  &  ^  a, 

and  (a  X  6)  -^  ft  =  a, 

42.  It  follows,  fi-om  the  definition  of  division,  that  if  the  product 
of  two  factors  be  divided  by  either  of  the  factors^  the  resulting  qvx)tient 
will  be  the  other  factor. 

Or,  (a  X  6)  -f-  a  =  6, 

and  {a  X  h)  -^  b  =  a. 

Since  in  the  product  (a  x  h)  the  factors  a  and  h  of  the  dividend  are  sep- 
arated by  the  miilti plication  sign,  it  is  merely  a  matter  of  inspection  to 
obtain  the  second  member  of  each  identity. 


POSITIVE  AND   NEGATIVE   NUMBERS  69 

43.  The  Law  of  Quality  Signs  for  division  may  be  obtained 
directly  from  the  set  of  identities  in  §  10  by  applying  the  definition 
of  division. 

(i.)     +(ab)  ^  +b  =  +a,  (iii.)   -(ab)  -^  +b  =  -a, 

(ii.)    +{nb)  ~-b=    a,  (iv.)    -{ab)  ^ -b  =  +a. 

It  should  be  observed  that  the  quotient  is  positive  whenever  the  signs  of 
quality  of  the  dividend  and  divisor  are  like,  as  in  (i.)  and  (iv.),  and  the 
quotient  is  negative  whenever  the  signs  of  quality  of  the  dividend  and 
divisor  are  unlike,  as  in  (ii.)  and  (iii.)* 

44.  It  follows  that  the  quotient  obtained  by  dividing  any  number 
by  "^1  is  equal  to  the  number  itself  It  follows,  also,  that  the  quo- 
tient obtained  by  dividing  any  number  by  ~1  is  a  number  equal  in 
absolute  value  to  the  dividend  but  opposite  in  quality. 

E.g. 


+5  -f  +1  =  +5, 

-7^ 

-+1^-7, 

+6  -f  -1  =  -6, 

-8-i 

-  -1  =  +8- 

45.  The  quotient  obtained  by  dividing  1  by  any  number  is  called 
the  reciprocal  of  the  number. 

E.  g.   The  reciprocal  of  5  is  ^. 

46.  Since  the  product  of  any  number  multiplied  by  its  reciprocal 
is  by  definition  "^1,  it  follows  that  any  number  and  its  reciprocal 
have  the  same  quality. 

E.  g.  The  numbers  ~3  and  ^^  are  reciprocals,  and  both  are  negative 
numbers. 

47.  Dividing  hy  any  number^  except  0,  produces  the  same  result 
as  multiplying  by  the  reciprocal  of  that  number. 

Representing  any  number  by  A,  and  any  other  number  different  from  0 
by  dy  we  maj''  represent  the  product  of  A  and  the  reciprocal  of  d  by  writing 
Ax{\^d). 

If  this  expression  be  multiplied  by  rf,  the  result  is  A. 

Hence,  A  x  (1  -f  rf)  is  equal  to  the  quotient  A  -^  d^  that  is, 

E.  g.    The  quotient  12  -^  3  is  equal  to  the  product  12  x  -• 


60  FIRST  COURSE  IN  ALGEBRA 

48.  As  an  extended  defiuition  of  division,  to  correspond  to 
that  of  multiplication,  we  have  the  following  : 

To  dimds  one  number  by  another  is  to  do  to  the  first  that  which 
must  be  done  to  the  second  to  obtain  the  positive  unit  "^1. 

49.  In  the  division  of  positive  and  negative  numbers,  we  may 
have 

I.  The  Divisor  Positive 

Ex.  1.   Divide  +24  by  +6. 

By  tlie  extended  definition  of  division,  the  quotient  resulting  from  the 
division  of  +24  by  +6  may  be  obtained  by  performing  upon  the  dividend 
+24  such  operations  as  must  be  performed  upon  the  divisor  +6  to  obtain  the 
unit  of  positive  numbers  +1. 

Since  +6  =  +1  x  6,  it  appears  that  we  may  obtain  the  unit  of  positive 
numbers  from  +6  by  dividing  the  absolute  value  of  +6  by  6. 

Hence  the  quotient  of  +24  -^  +6,  is  a  positive  number  obtained  by  divid- 
ing the  absolute  value  of  +24  by  6  ;  that  is  : 

+24  -^  +6  =  +(24  -f  6)  =  +4. 

Ex.  2.    Divide  -30  by  +10. 

Reasoning  as  before,  the  quotient  will  be  the  negative  number  obtained 
by  dividing  the  absolute  value  of  the  dividend  "30  by  10. 

That  is,    30  -^  +10  =  -(30  ^  10)  =  -3. 

II.  The  Divisor  Negative 

Ex.  3.   Divide  +32  by  -16. 

By  the  extended  definition  of  division,  we  may  obtain  the  quotient 
resulting  from  the  division  of  +32  by  "16  by  treating  the  dividend  +32  in 
the  same  way  as  we  treat  the  divisor  ~16  to  obtain  the  unit  of  positive 
numbers  +1. 

By  first  reversing  the  quality  of  "16  we  may,  from  the  positive  number 
thus  obtained,  +16,  obtain  the  unit  of  positive  numbers  +1,  by  dividing  the 
absolute  value  of  the  result  by  16. 

Hence  we  may  obtain  the  desired  quotient  by  first  reversing  the  quality 
of  the  dividend  +32,  and  dividing  the  absolute  value  of  the  result  thus 
obtained  by  16. 

That  is,'  +32  -^  -16  =  "(32  ^  16)  =  "2. 

Ex.  4.    Divide  "40  by  "8. 

In  order  to  obtain  the  unit  of  positive  numbers  +1  from  the  divisor  "8, 
we  may  first  reverse  the  quality  of  the  divisor,  obtaining  a  positive  number 
+8,  and  then  divide  the  absolute  value  of  the  number  thus  obtained  by  8. 

Hence  we  may  perform  the  same  steps  with  respect  to  the  dividend  "40. 

That  is,  -40  -f  "8  =  +(40  -f  8)  =  +5. 


POSITIVE  AND   NEGATIVE   NUMBERS  61 


Exercise  V.   S 

; 

Simplify  the  following 

■: 

1.    +6 --+2. 

11.   -15 -f- -3. 

21.   -42-^+14. 

2.    +9-^+3. 

12.  -l2-^+4. 

22.    +50  -T-  -1. 

3.    +10-^+5. 

13.   -6-f-+6. 

23.    +56 -f- -7. 

4.    +12^+6. 

14.    +17-^-17. 

24.   -57  -^  -19. 

5.    +14 -h +2. 

15.   -19-r--19. 

25.    +63-^-9. 

6.    +16^-4. 

16.    +13-^-13. 

26.    -64-^-16. 

7.    +18-^-9. 

17.    +27  -^  -9. 

27.    +65-7- -13. 

8.    +20 -f- -5. 

18.    -33  — +3 

28.    "68  -i-  +17. 

9.    -8^+4. 

19.   -38-f--19. 

29.    +70H--14. 

10.   -4-H+2. 

20.    +40-^+10. 

30.   -75 -f- +5. 

31.    (+36  X 

-2) 

-T-  -8.              35. 

(+56 

-f-  -8)  X  -7. 

32.    (+15  X 

-4) 

-r-  -12.           36. 

(+32 

-^  -16)  X  -5. 

33.    (-24  X 

-3) 

4-  +9.              37. 

(-24 

-^  -2)  -T-  "2. 

34.    (+16  X 

+4) 

4-  -8.             38. 

(+48 

-r-  -12)  ^  -4. 

Find  the  value  oi  a  -^  (b  +  c)  when 
39.    a  =  -27,  6  =  +5,  c  =  +4.  40.   a  =  -39,  b  =  +15,  c  =  '2. 

Find  the  value  oi  (a  +  b)  -^  (c  +  (T)  when 
41.    a  =  +l,b=+2,c  =  +4:,d=-L     42.   a==+ll,^>=-2,  c  =  +6,  ^=+3. 

50.  Commutative  Law  for  Division.  Since  multiplications 
may  be  performed  in  any  order,  it  follows  that,  in  a  series  of  succes- 
sive divisions  also,  the  operations  may  be  performed  in  any  order ; 
that  is, 

51.  In  any  chain  of  operations  containing  both  multiplications 
and  divisions,  the  quantities  may  be  rearranged  in  any  order,  provid- 
ing the  sign  of  operation,  X  or  -f-,  attached  to  any  particular  operand, 
moves  with  it  when  it  changes  from  one  position  to  another. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Representing  arithmetic  whole  numbers  by  a,  6,  and  c,  consider  axb-^c. 

By  the  principle  of  §  41  it  follows  that,  if  any  number  a  be  divided  by 
any  number  c,  except  zero,  and  this  result  be  then  multiplied  by  c,  the  result 
will  be  the  same  as  if  no  operation  had  been  performed  upon  a.     That  is  : 

a  -^  c  X  c  =  a. 


t)ii  FIRST   COURSE  IN  ALGEBRA 

Substituting  this  expression  for  a  in  the  given  expression,  we  nia}'  write 

ff  X  6  -^  c  =  (a  -r  c  X  f)  X  i  -^  ('. 
Applying  the  Commutative  Law  for  multiplication  to  tlie  two  factors 
c  and  bj  we  have  : 

{a  -7-  c  X  c)  x6-7-c  =  a-rCX&xc-^c. 

In  this  chain  we  may  neglect  x  c  ^  c  as  producing  no  change  in  the  final 
result.     Therefore  axb-^c  =  a-^cxb. 

It  foUowsi  that^  in  an  unbroken  chain  of  multiplications  and  divisions,  the 
operations  may  be  performed  in  any  ^order. 

E.g.  2-^7  x  14  =  2  X  14^7  =  28-r7  =  4. 

52.  It  follows,  from  the  Law  of  Commutation  for  multiplications 
and  divisions  occurring  together,  that  a  pmdact  of  two  w  more 
factors  may  he  dicided  hi/  a  number  by  dividing  one  of  the  factors 
of  the  jwoduct  by  tlmt  number. 

(The  f ollowiug  proof  may  be  omitted  when  the  cliapter  is  read  for  the  first  time. ) 

Representing  any  positive  integral  numbers  by  «,  6,  and  c,  c  not  being  0, 
we  have  the  followinjj  : 


(a6)  -fc  =  ax6-rC, 
=  rt  -^  c  X  6| 

=  (rt  -r  c)  X  6; 


{alj)  -=-c  =  ax6-rC,  Notation 

=  b  -^  c  X  a^  Commutative  Law 
=  (6  -^  c)  X  «,  Notation 
=  rt  X  (?>  -r  c),  Commutative  Law 
From  the  reasoning  above  it  follows  that 

(ofc)  ^  ^  =  f  either  («  ^  c)  X  6 
I  or        a  X  (&  -f  c). 

Associative  Laiir  for  Division 

Principles  Governing  the  Removal  and  Insertion  of 
Parentheses 

63.  Parentheses  preceded  by  tlie  sign  of  multiplication  X. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 
Representing  arithmetic  values  of  whole  numbers  by  a,  b,  and  c,  as  in 

similar  cjises,  c  not  being  0,  consider  the  expression  a  x  (b  -^  c)  in  which 

parentheses  are  preceded  by  a  multiplication  sign. 

Since  successive  multiplications  and  divisions  by  c  produce  no  change  of 

value  in  the  result,  we  may  write 

ax  (b  -^  c)  =  a  x  (b  ^  c)  x  c  -r  c. 

By  the  definition  of  division    (b  -^  c)  x  c  =  b. 

Hence  ax  (b  -^  c)  xc-^c  =  axb-^c. 

Therefore  a  x  (b  -^  c)  =  a  x  b  -^  c. 


POSITIVE  AND  NEGATIVE   NUMBERS  63 

An  unbroken  chain  of  multiplications  and  divisions  may  he  re- 
moved from  parentheses  preceded  by  the  sign  of  multiplication  with- 
out altering  the  signs  of  multiplication  and  division  preceding  the 
different  numbers  removed.     (Compare  with  Prin.  I.  (i.)  Chap.  IV. 

§'7.) 

In  case  either  the  sign  of  multiplication  or  the  sign  of  division  is 
required  before  the  first  number  enclosed  within  parentheses, 
and  neither  sign  is  written,  the  sign  of  multiplication  is  to  be 
understood. 

54.  Since  the  proof  of  any  identity  establishes  also  the  truth  of 
its  converse,  we  may  state  that  an  unbroken  chain  of  multiplications 
and  divisions  may  be  enclosed  ivithin  parentheses,  preceded  by  the 
symbol  of  multiplication,  without  altering  the  signs  of  multiplication 
and  division  attached  to  the  numbers  included.  (Compare  with 
Prin.  II.  (i.)  Chap.  IV.  §  8.) 

55.  Parentheses  Preceded  by  the  Sign  of  Division  -f-. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time. ) 

Consider  the  expression  a  -f  (6  -f  c)  in  which  parentheses  are  preceded 
l)y  the  sign  of  division. 

It  may  be  seen  that  the  chain  of  successive  operations  represented 
by  X  6  -f  c  X  c  -f  b,  if  applied  to  any  number,  will  not  affect  its  value. 
Hence,  we  may  write 

a-r.  (6-rc)  =  «-^(^-^c)x&-^cxc-^6 

=  «  -f  (6  -f  c)  X  (6  -^  c)  X  c  -^  &. 

Wc  may  neglect  the  successive  operations  represented  l)y  -f-  (h  -f  c) 
X  (h  -^  c),  as  producing  no  alteration  in  the  value  of  a. 

Hence,  «  -f  (6  -f  c)  x  (h  -^  c)  x  c  -^  h  =  a  x  c  -^  b. 

Applying  the  Commutative  Law  for  multiplications  and  divisions,  we 
may  write  finally 

a  -^  (b  -^  c)  =  a  -^  h  X  c. 

An  unbroken  chain  of  multiplications  and  divisions  may  be  re- 
moved from  parentheses  preceded  by  the  sign  of  division,  providing 
the  signs  of  multiplication  and  division  preceding  the  different  num- 
bers removed  be  changed  from  X  to  -^,  or  from  -^  to  X.  (Compare 
with  Prin.  I.  (ii.)  Chap.  IV.  §  7.) 

56.  Since  the  proof  of  any  identity  establishes  also  the  truth  of 


64  FIRST  COURSE  IN  ALGEBRA 

its  converse,  it  follows  that  an  uvibrohen  chain  of  multiplications 
and  divisions  may  be  enclosed  within  parentheses  preceded  by  the 
symbol  of  operation  for  division,  provided  the  symbols  of  operation 
for  multiplication  a  fid  division,  attached  to  all  numbers  enclosed,  be 
reversed,  from  X  to  -^  or  from  -7-  to  X.  (Compare  with  Prin.  II.  (ii.) 
Chap.  IV.  §  8.) 


E.  g.   Parentheses  preceded  by  x 

+6  X  (-2  -r  -3)  =  +6  X  -2  ^  -3 
=  -12  -^  -3 

=  +4. 


Parentheses  preceded  by  ~- 

+32  -f  (-8  -^  -2)  =  +32  -^  -8  X  -2 
=  -4  X  -2 

=  +8. 


Observe  that  +32  -^  ("8  -^  "2)  is  not  equal  to  +32  -^  "8  -f  "2.  For,  we 
have  +32  -f  "8  4-  ~2  =  "4  -f  "2  =  +2. 

57.  From  the  Associative  Law  for  multiplications  and  divisions 
we  have  the  following  Law  of  Signs : 

(i.)         X  (X  a)  =  X  a,  (iii.)        -^  (X  a)  =  -i-  a, 

(ii.)      -^  (-f-  a)  =  X  a,  (iv.)         X(-^a)  =  -^a, 

It  appears  that /or  two  like  signs  either  of  multiplication  or  of 
division,  occurring  successively  as  in  (i.)  and  (ii.),  we  may  substitute 
th£  direct  sign  X  ;  and  for  two  unlike  signs  occurring  successively,  as 
in  (iii.)  and  (iv.),  we  may  substitute  the  indirect  sign  -i-.  (Compare 
with  §  10.) 

Exercise  V.    4 

Simplify  the  following  arithmetic  expressions  : 

1.  4  X  (3  -r-  2).  7.    12  -H  (e  -h  3  X  2). 

2.  1  -^  (1  ^  6).  8.    15  X  (3  -h  5  X  4). 

3.  1-T-(2X11).  9.    14^(7 -^  5 -^  3). 

4.  5  ^  (4  -^  5).  10.    5  -^  [4  -f-  5  X  (3  -^  5)]. 

5.  27  -T-  (9  -T-  2).  11.    7  -r-  [8  X  3  -^  (12  -f-  7)]. 

6.  21  ^  (7  -^  4).  12.    16  X  [9  -^  8  X  (1  -^  3)]. 

Simplify  the  following  algebraic  expressions  : 

13.  +10  X  (+3  -^  +2).  17.    +1  -H  (+6  -T-  -1). 

14.  +28  X  (-4  -^  -7).  18.   -16  -^  (+8  -f-  +4). 

15.  +9  X  (+8  4-  -3).  19.   -27  -^  (-8  -^  +3). 

16.  +3  -T-  (+4  -^  +5).  20.   -1  -^  (-1  -f-  -2). 


POSITIVE  AND  NEGATIVE  NUMBERS  65 

Distributive  Law  for  Division. 

58.  Ill  division  the  dividend  may  be  distributed,  the  signs  of  the 
partial  quotients  following  the  same  law  of  signs  as  the  partial 
products  in  multiplication.     Hence,  the  following 

Principle:  The  quotient  resulting  from  the  division  of  any  alge- 
braic sum  by  a  single  algebraic  number  may  be  obtained  by  dividing 
each  term  of  the  dividend  by  the  divisor.  The  signs  prefixed  to  the 
partial  quotients  thus  obtained  are  -{•  or  —  according  as  the  terms 
from  which  they  are  obtained  by  division  have  like  or  unlike  quality 
signs. 

{The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

As  in  similar  cases  let  a,  />,  <?,  and  d  represent  any  positive  integral  values, 
with  the  restriction  tliat  d  shall  not  be  0.     Consider  {a  —  h  -\-  c)  -^  d. 
By  the  definition  of  division,  we  may  write 
-{-  a  —  d  X  d  =  -\-  a, 
—  h-^dx  d  =  —  bf 
+  c-^dxd  =  +  c. 
Substituting  these  values  for  a,  b  and  c,  in  the  expression  a  —  b  +  c,we 
may  write 

(a  —  b  -\-  c)  -^  d  =  [a  -^  d  X  d  —  b  -^d  X  d  +  c  -^  d  X  d]-^  d. 
Observe  that  the  expression   in   square  brackets   may  be   obtained   by 
multiplying  the  following  expression  by  d: 

a-^d  —  b-^d  +  c-^d. 
Hence^  (a  —  b  +  c)  ~  d  =  [(a  -i-d  —  h^d-\-c-!^  d)]  x  d  -^  d 

=  a  -^  d  —  b  -^  d  +  c  -^  d. . 
E.  g.  [+12  X  -26  -  +8  X  +13  +  -11  X  "39]  -f  +13 

=  [+12  X  -2  -  +8  X  +1  +  -11  X  -3] 
=  -24  +  -8  +  +33 
=  +1. 

59.  Although  the  dividend  may  be  distributed,  the  divisor 
cannot  be. 

E.  g.    It  should  be  observed  that     :  = =  H v ' 

'='  c-^-dc  +  dc  +  d 

but    —f-j  ^  — ^—  +  —^  . 
c  +  d/       c  d 

60.  Division  of  one  expression  by  another  may  be  indicated  by 
writing  the  dividend  as  the  numerator  and  the  divisor  as  the 
denominator   of  a  fraction.      When  either  the   dividend  or  the 

6 


QQ  FIRST  COURSE   IN  ALGEBRA 

divisor  consists  of  more  than  one  term,  the  horizontal  dividing  line 
of  the  fraction,  which  separates  numerator  from  denominator,  serves 
both  as  a  sign  of  division  and  of  grouping. 

E.g.  ^=^a  +  b)^(c-d). 

61.  In  a  chain  of  operations  involving  additions,  subtractions, 
multiplications  and  divisions,  unless  the  contrary  is  specified,  the 
multiplications  and  divisions  must  be  performed  first,  and  then  the 
indicated  additions  and  subtractions. 

However,  by  the  use  of  parentheses  we  may  indicate  that  addi- 
tions and  subtractions  are  to  be  performed  before  multiplications 
and  divisions. 

Ex.  1.  +2  X  +3  +  -4  X  -5  +  -8  -^  +2  =  +6  +  +20  +  "4  =  +22. 

Ex.  2.         +2  X  (+3  +  -4)(-5  +  "8  ^  +2)  =  +2  x  (-l)(-5  +  -4) 

=  -2  (-9)  =  +18. 

62.  From  our  definition  of  0  as  resulting  from  the  subtraction  of 
any  number  from  an  equal  number,  it  follows  that  the  quotient  ob- 
tained, by  dividing  zero  by  any  number  different  from  zero  is  zero. 
For,  letting  a  and  x  represent  any  numbers  which  are  different 
from  zero,  we  have  0~-x  =  {a  —  a)-^x  =  a~-a'  —  a-^£c  =  0. 

63.  It  follows,  conversely,  that  if  a  quotient  be  0  ths  dividend  must 
be  0,  since  from  the  nature  of  the  case  the  divisor  cannot  be  0. 

That  is,  b  being  any  number  other  than  0, 

.  if  a  -h  ^  =  0,  then  a  must  equal  0. 

Exercise  V.  5 

Simplify  the  following  expressions  : 

1.  +20  -^  -5  +  -30  -h  +6  -  -40  ^  -8. 

2.  -24  X  -2  -  +3  -^  -1  +  -9  X  +5. 

3.  +12  ^  -3  X  +4  -  -15  X  -5  -f-  +3  +  +1. 

4.  -36  X  -2  -^  +12  —  +1  -^-  +1  X  -2  —  "S. 

5.  +6  X  +5  +  -7  X  -2  +  +3  X  -4  -+-  -1  X  +9. 

6.  -8  -^  +2  —  +15  X  -1  -  -14  -^-  "7  +  "6  X  +2. 

7.  (+2  +  +6)  (-3  +  -4)  +  +50. 

8.  (-7  -  +5)(-l  +  -10  ^  +2). 

9.  +3-^(-10--ll)(+8--5-T--l). 

10.   (-2  X  +4  -  -4  -^  +2)  -^  (+6  -r-  -3  -  -1  -^  +1)  X  -7. 


POSITIVE  AND  NEGATIVE  NUMBERS  67 

64*  Tlie  Fundamental  Laws  of  ordinary  Algebra,  as  discussed  in  the 
preceding  pages,  have  been  proved  for  special  cases  only,  and  not  in  their 
most  general  forms. 

By  the  Principle  of  No  Exception  they  will  now  be  assumed  to  be 
extended  to  include  all  numbers  which  are  positive  or  negative,  integral  or 
fractional,  and  will  be  laid  down  as  being  fundamental  to  the  science  of 
Algebra. 

Our  problem  will  now  be  to  so  define  and  interpret  our  symbols  and  the 
results  of  operations  with  them,  that  they  shall  be  consistent  with  the  Com- 
mutative, Associative,  and  Distributive  Laws,  which  are  three  fundamental 
laws  of  Algebra. 

65.*  By  means  of  the  Law  of  Commutation  we  have  obtained  the 
principles  governing  the  insertion  and  removal  of  parentheses,  and  by  the 
aid  of  these  principles  we  have  developed  the  Law  of  Distribution. 

By  applying  these  principles  repeatedly  it  may  be  shown  that  the  Dis- 
tributive Law  for  Multiplication  holds  when  both  multiplicand  and  mul- 
tiplier consist  of  more  than  one  number  or  term  ;  that  is,  with  the  restriction 
that  a  >  6  and  c  >  dy  we  may  show  that 

(a  —  b)(c  —  d)  =ac  —  be  —  ad-\-  bd. 

By  letting  a  and  c  each  equal  zero,  we  obtain  from  the  above, 
(-bX-d)=  +  bd, 
from  which  \ve  obtain  the  following  Law  of  Quality  Signs  : 
(+-hX+-d)  =  ++bd. 

The  remaining  forms  for  the  Law  of  Signs  for  different  combinations  of 
signs  of  operation  and  of  quality  (see  §  10)  may  be  shown  to  hold  true  by  a 
similar  course  of  reasoning. 

We  thus  obtain  the  Law  of  Quality  Signs  as  a  direct  result  of  the  Law 
of  Distribution  for  Multiplication,  and  accordingly  the  Law  of  Signs  may 
be  included  among  the  fundamental  laws  of  operation. 

66.  We  shall,  in  the  following  chapters,  employ  a  single  set  of 
signs,  +  and  — ,  to  denote  both  the  operations  of  addition  and 
subtraction,  and  the  qualities  of  the  numbers  to  which  they  are 
attached,  as  being  positive  or  negative. 

The  symbol  of  operation  +  preceding  a  number  or  letter  which 
stands  first  in  a  chain  of  additions  and  subtractions  will  usually  be 
omitted,  whether  it  denotes  the  operation  of  addition  or  the  quality 
of  the  number  as  being  positive ;  but  the  sign  — ,  whether  indicating 

*  Thifl  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


68  FIRST  COURSE   IN  ALGEBRA 

the  operation  of  subtraction  or  the  negative  quality  of  the  number  to 
which  it  is  attached,  can  never  be  omitted. 
In  expressions  such  as  the  following 

(+3)  -  (-4)  +  (-5)  -  (-G) 

each  sign  within  parentheses  is  to  be  interpreted  as  indicating 
quality,  and  each  sign  outside  of  the  parentheses  as  indicating  an 
operation. 


ADDITION   AND   SUBTRACTION 


CHAPTER  VI 

ADDITION  AND  SUBTRACTION  OF  INTEGRAL  ALGEBRAIC 
EXPRESSIONS 

Definitions 

1.  An  algebraic  expression  may  for  the  present  be  defined  to  be 
any  collection  or  combination  of  letters  or  of  letters  and  numbers,  con- 
nected by  the  signs  of  operation  +,  — ,  X  and  -r-,  which  may  be  used 
according  to  the  principles  and  definitions  of  algebra  to  represent  a 
number.     (Compare  with  Chap.  I.  §12.) 

2.  The  parts  of  an  algebraic  expression  which  are  separated  by 
the  signs  plus  and  minus  are  called  the  terms. 

3.  Whenever  any  term  is  regarded  as  being  separated  into  two 
factors,  either  factor  may  be  called  the  co-factor  or  the  coeflacient 
of  the  other. 

E.  g.  In  the  term  5  abed,  5  is  the  coefficient  of  ahcd,  5  a  of  bed,  ac  of  5  bd, 
etc. 

Thus  in  Axyz,  4  is  the  coefficient  of  xyz;  ^x  that  of  yz  ;  and  4xy  that 
of  z. 

When  one  of  the  factors  of  a  product  is  a  numeral  symbol,  it  is 
called  the  numerical  coefficient  of  the  product  of  the  other 
factors. 

Unless  the  contrary  is  specified,  when  we  speak  of  the  coefficient 
of  a  term  we  mean  the  numerical  coefficient  taken  together  with  the 
sign  +  or  —  preceding  it.  When  no  numerical  coefficient  is  written, 
unity  is  understood. 

4.  A  power  of  a  number  is  a  product  obtained  by  using  that 
number  two  or  more  times  as  a  factor. 

Thus  the  second  power  of  2  is  2  x  2  or  4 ;  the  third  power  is  2  x  2  x  2 
or  8  ;  the  fifth  power  is  2  x  2  x  2  x  2  x  2  or  32. 

5.  For  the  present  we  shall  define  an  exponent  as  an  integral 
number  written  at  the  right  of  and  a  little  above  a  number  or 


70  FIRST  COURSE  IN  ALGEBRA 

expression  to  show  how  many  times  the  number  or  expression  is  to  bo 
taken  as  a  factor. 

Thus  a\  read  "a  square"  or  "a  to  the  second  power,  "  means  a  >  a\a^, 
read  "a  cube"  or  "a  to  the  third  power,"  means  a-a-a^  afi  means 
a.a-a.a.a.a-y  etc. 

a;**  may  be  interpreted  as  meaning  a  number  x  taken  as  a  factor  n  times, 
and  not  until  definite  values  are  assigned  to  x  and  n  will  the  expression  be 
regarded  as  having  a  definite  numei'ical  value.  (See  also  Chap.  XIX 
§§1-4.) 

The  same  notation  may  be  applied  to  expressions  in  which  two  or 
more  numbers  or  letters  appear. 

E.  g.  («&)*  means  {ah){ah'){(ih){ah)  -, 

(x  +  yY  means  \x  +  y){x  +  y){x  +  y){x  +  y) ; 
(5  -  zY  means  (5  -  2!)(5  -  z){b  -  z). 

When  no  exponent  is  written,  the  exponent  1  is  understood. 

Thus,  2  is  called  the  first  power  of  2  ;  5  the  first  power  of  5  ;  a  the  first 
power  of  a ;  the  exponent  1  being  understood  in  each  case. 

6.  The  expression  "  no  exponent "  must  not  be  confused  with 
the  "  exponent  zero.  "  We  shall  show  later  that  any  number  with 
the  exponent  zero  may  be  regarded  as  standing  for  unity  or  1. 

7.  One  power  is  higher  or  lower  than  another  according  as  its 
exponent  is  greater  than  or  less  than  that  of  the  other. 

E.  g.  The  fourth  power  of  a  number  is  "  higher  "  than  the  second  power  ; 
the  sixth  is  higher  than  the  fifth  ;  etc. 

8.  A  power  is  even  or  odd  according  as  its  exponent  is  even  or 
odd. 

9.  Any  letter  or  number  which  is  raised  to  a  power,  is  called  a 
base. 

10.  Each  literal  factor  of  a  term  is  called  a  dimensioR  of  the 
term. 

It  is  customary  to  write  the  literal  factors  of  a  term  in  alphabetical 
order,  unless  for  some  particular  reason  a  different  arrangement  is 
required. 

11.  The  number  obtained  by  adding  the  exponents  of  the  literal 
factors  of  a  term  is  called  the  decree  of  the  term  as  a  whole. 


I 


ADDITION  AND  SUBTRACTION  71 

E.  g.  y^,  x^ij,  abc,  are  of  the  third  degree  ; 

abcde^  x^  x^y,  m^n'^,  are  of  the  fifth  degree ;  etc. 

12.  Similar  or  like  terms  are  terms  which  contain  the  same 
letters,  affected  by  the  same  exponents.  These  terms  may,  however, 
differ  in  their  numerical  coefficients. 

Thus,  5  xy  and  —  10  xy  are  like  terms ;  so  also  are  7  a%h  and  9  a%h. 

13.  An  algebraic  expression  is  called  a  monomial,  binomial, 
or  trinomial,  according  as  it  consists  of  one,  two,  or  three  terms. 
Algebraic  expressions  of  two  or  more  terms  are  commonly  spoken  of 
as  polynomials  or  as  multinomials* 

14.  A  monomial  is  integral  if  the  letters  which  it  contains 
enter  by  multiplication  only,  and  none  enter  by  division,  that  is, 
if  none  appear  in  the  denominator  of  any  fraction. 

E.  g.  abCf  mn\  2  xy^  15  xy^z^^  are  all  monomials. 

aH-&,  2a:  —  3 1/,  -T-+1  are  binomials. 

a;  +  y  +  s,  aa;2  +  ?>a:  +  c,  2  m  +  3  a;  -  11,  are  trinomials, 
a*  +  a*  +  a'*  +  «  4-  1  is  a  polynomial. 

All  of  these  expressions,  with  the  exception  of  the  monomials,  are  poly- 
nomials or  multinomials. 

15.  A  polynomial  is  said  to  be  integral  with  respect  to  a 

specified  letter  when  this  letter  does  not  appear  in  the  denominator 

of  any  fraction  ;  that  is,  when  it  does  not  enter  any  term  through  the 

process  of  division.     In  the  opposite  case  it  is  said  to  be  fractional. 

,       a2  _  52             (^  _  ly           a'^  +  -2ah  +  h'^  . 
E.  g.    The  expression  x^  4 r —  x^  —  - — —-tto  x  +  is 

integral  with  respect  to  «,  but  fractional  with  respect  to  a  and  h. 

In  order  that  a  polynomial  be  integral  with  respect  to  a  given 
letter  it  is  necessary  that  all  of  the  terms  which  appear  in  it  be  inte- 
gral with  respect  to  that  letter. 

E.  g.  a:*  +  5  a;2  —  x  -f  1  is  an  integral  polynomial  of  the  third  degree  with 
reference  to  x,  since  the  highest  power  of  x  which  appears  is  the  third. 

We  may  regard  ao"  +  fex^  +  ex  +  (i  as  a  polynomial  of  the  third  degree 
with  reference  to  x  alone,  or  of  the  fourth  degree  if  no  particular  letter  is 
specified.  This  is  because  the  term  ax*  is  of  the  third  degree  with  reference 
to  X  alone,  but  of  the  fourth  degree  with  reference  to  a  and  x  taken  together. 


72  FIRST  COURSE   IN  ALGEBRA 

16.  An  expression  in  which  all  of  the  terms  are  of  the  same  degree, 
reckoned  with  reference  to  all  of  the  letters,  is  hoino^eneous. 

E.  g.  X*  +  x^y  +  x^y^  +  xy^  +  y^  and  abc  +  5  a'  +  bh  are  homogeneous, 
and  of  the  fourth  and  third  degrees  respectively. 

17.  One  of  the  equal  factors  of  a  number  is  called  a  root  of  the 
number.  According  as  there  are  two,  three,  or  four  equal  factors, 
etc.,  each  is  called  a  square  root,  cube  root,  fourth  root,  etc. 

Thus,  2  is  one  of  the  square  roots  of  4  ;  3  of  9 ;  7  of  49 ;  etc.  2  is  a  cube 
root  of  8 ;  3  of  27  ;  5  of  125  ;  etc. 

18.  A  root  of  a  number  is  indicated  commonly  by  means  of  a 
root  or  radical  sign  ^/~, 

A  small  number,  called  the  index  of  the  root,  placed  thus, 
/^~,  >y/"~,  '^~,  etc.,  is  used  to  indicate  the  order  of  the  root,  that 
is,  whether  it  be  a  second,  third,  or  fourth  root,  etc.  If  no  index  be 
expressed,  the  index  2  is  understood. 

Thus,  ^^27  is  3,  'C^  is  2,  {/32  is  2,  and  ^/'W  =  ^IQ  is  4. 

19.  An  expression  is  rational  with  respect  to  specified  letters 
when  it  does  not  contain  indicated  roots  of  these  letters.  In  the 
contrary  case,  it  is  said  to  be  irrational. 

According  to  this  definition,  a  rational  expression  may  contain 
indicated  roots  of  the  numerical  parts. 

■c         o    2      m  +  n     V5  ahc    +  3  a  +  V'4  6  .     . 

E.  g.    2  xhjz,  — ,  — >  -—^ ,  are  rational  with  respect 

o  ft  9  xy  2i 

to  the  letters,  for  no  letter  appears  under  a  radical  sign.     On  the  other 

. 2  a \/a^  +  y^ 

hand,    v  a;^  —  y^  and are  irrational   with   respect  to  the 

Va  —  b 
letters. 

20.  A  polynomial  is  rational,  irrational,  integral,  or  frac- 
tional according  as  its  terms  are  rational,  irrational,  integral,  or 
fractional. 

21.  A  polynomial  is  said  to  be  arranged  with  reference  to  a 
specified  letter  when  the  exponent  of  that  letter  in  successive  terms 
increases  or  decreases  in  numerical  value. 


ADDITION   AND   SUBTRACTION  73 

E.  g  The  expression  4  +  2  a;  +  6  x^  -f  5  a;^  +  a;*  is  arranged  according  to 
increasing  powers  of  a;;  in  the  following  form  it  is  arranged  according  to 
decreasing  powers  of  x  ;  a,-^  +  5  x^  +  6  a;^  +  2  a:  +  4. 

The  polynomial  a^  +  5  a%  +  10  a%^  +  10  aW  +  5  ah*  +  h^  is  arranged 
according  to  descending  powers,  5,  4,  3,  2,  1,  of  a,  and  according  to  the 
ascending  powers,  1,  2,  3,  4,  5,  of  b. 

Addition  of  Monomials 

22.  The  algebraic  sum  of  two  numbers  or  quantities  may  always 
be  indicated  by  writing  them  one  after  the  other,  separated  by  the 
sign  of  addition,  each  term  being  preceded  by  the  sign  of  quality  of 
its  numerical  coefficient. 

This  collecting  of  several  numbers  or  quantities  into  one  algebraic 
expression  is  what  in  algebra  is  called  addition.  The  resulting 
expression  is  called  the  sum. 

23.  It  is  commonly  understood  that,  after  the  parts  of  a  sum  are 
written  consecutively  as  parts  of  one  algebraic  expression,  like  or 
similar  terms  are  to  be  united. 

This  "reduction  "  is  not  addition,  but  amounts  simply  to  changing 
the  form  of  an  expression  so  that  it  shall  have  as  few  terms  as 
possible. 

(a)   Addition  of  Dissimilar  Terms. 

24.  If  several  quantities  or  terms  to  be  added  are  unlike,  they 
cannot  he  united  to  form  any  particular  amount  or  number  of  either. 
TJw  sum  may  be  indicated  by  writing  the  terms  one  after  the  other 
with  the  proper  signs. 

E.  g.  The  sum  of  a  and  h  may  be  indicated  by  writing  a  +  6,  but  not 
until  particular  values  are  given  to  a  and  h  can  we  find  a  single  7iumber 
representing  their  sum. 

If  a  represents  2  and  6  represents  5,  then  ct  +  6  represents  7. 

If  a  represents  +  3  and  h  represents  —  3,  then  a  +  6  represents  0. 

Ex.  1.    Express  the  sum  of  3x  and  —  2i/. 

Since  these  terms  are  unlike,  we  cannot  combine  them  into  a  single  term, 
but  we  may  express  the  sum  by  writing  the  terms  separated  by  a  positive 
sign,  considering  the  second  term  as  a  negative  number  ;  that  is 

3x+(-27/)  =  3a;-2i/. 


74  FIRST   COURSE   IN   ALGEBRA 

(b)  Addition  of  Similar  Terms. 

25.  Like  m-  similar  terms  may  be  united  by  addition  into  a  single 
term. 

Ex.  2.   Find  the  sura  of  5  a&,  3  ah,  and  ah. 

Since  we  have  no  knowledge  of  the  value  represented  by  the  product  of 
the  letters  a  and  6,  we  may  add  the  terms  with  respect  to  the  product  ah,  as 
a  concrete  number. 

Indicating  the  sum  by  writing  5  a&  +  3  a5  +  ah,  we  find  that  we  have 
(5  +  3  +  1)  a6,  that  is,  9  ah. 

Hence  we  may  write  5  aft  +  3  a6  +  a&  =  9  a6. 

Ex.  3.    Find  the  sum  of  4  mji*,  —  2  mii^,  and  —  5  mn^. 

Observe  that  the  terms  are  simihir  with  respect  to  the  literal  parts,  mn^. 
Hence  we  may  indicate  the  sum  by  writing  as  a  coefficient  to  this  common 
literal  part  the  algebraic  sum  of  the  numerical  coefficients  considered  as 
positive  and  negative  numbers  ;  that  is,  we  may  write 

4  mn^  —  2  mn^  —  5  mn^  =  (4  —  2  —  5)  min^  =  —  3  mn^. 

26.  The  addition  of  terms  in  algebra  is  performed  according  to 
the  following  Principles : 

(i.)  The  sum  of  two  monomials  may  be  indicated  by  writing  them 
with  their  signs  of  quality  one  after  the  other,  separated  by  the  sign  +. 

(ii.)  To  add  like  terms,  find  the  algebraic  sum  of  their  coefficients 
considered  as  positive  or  negative  numbers,  and  prefix  this  as  a 
coefficient  to  their  common  parts. 

Mental  Exercise  VI.    1 
Perform  the  following  indicated  additions: 


1. 

a 

2. 

-b 

3. 

2c 

4. 

5d 

a 

-b 

3c 

d 

5. 

-^9 

6. 

6A 

7. 

hx 

8. 

—  Ix 

-     9 

4A 

-2x 

\\x 

9. 

-Sx 

10. 

—  12y 

11. 

dz 

12. 

nw 

^x 

-133/ 

-Idz 

—  20w 

.3. 

Sab 

14. 

6  6c 

15. 

—    Zad 

16. 

Uxy 

2ab 

9  6c 

—  Wad 

—  9xy 

.7. 

4:C^X 

18. 

^by^ 

19. 

-llcW 

20. 

-26^y 

h^x 

8V 

21  c^" 

31^/ 

ADDITION  AND  SUBTRACTION  T5 


21. 

ax 
2  ax 
Sax 

22. 

5bi/ 
2by 

23. 

6c^ 
2^ 
5c^ 

24. 

1  dw 

—  5dw 

3dw 

25. 

A:£CW 

—  2xw 

—  xw 

26. 
30. 
34. 

11/ 

-3if 

3/ 

—  13  ^>a-^ 

—  Ibxij 

—  dbxij 

2S  ab\ 

—  ab'c 

—  5  ab^c 

27. 
31. 
35. 

12  z^w 

4:Z^W 
—  4:Z^W 

28. 
32. 
36. 

-2{)w^ 
10  w"" 

-    5w'' 

29. 

—  Sabc 

—  5  abc 

—  7  abc 

22a^bc 

a^bc 

-Sa%c 

Ucm^ 
18  cm^ 

-lldk^ 

-  13  dh^ 

2ddh^ 

33. 

Slxj/z' 

—  llxyz^ 

—  xyz^ 

-43a;y 

-27a;y 

xY 

37. 

la% 
5a'b 
Sa'b 
^a% 

38. 

dbc 

-10  be 

12  be 

—  Ubc 

39. 

-  5x^y 
dx'y 

-  13x^y 
ISx^y 

40. 

Ixyz 

—  8xyz 

—  dxyz 
10  xyz 

41. 

32  ab 
lab 

Uab 
2ab 

42. 

13a;y 
15  X1J 
llxij 
l^xy 

43. 

18  xhjz^ 
12  xhjz^ 
UxSyz"" 
Ux^yz^ 

44. 

UxYz^ 
UxYz^ 
22  xY^^ 
25a;y5« 

45. 

13  abx 

—  15  abx 

21  abx 

12  abx 

46. 

-23b\H 
37  b\H 

-  ^b\H 

-  b\H 

47. 

—  25  a^mw 

—  35  a^mw 

—  45  a^mw 
95  a^mw 

48. 

8  7wwaj^ 

11  mno? 

llmnx^ 

—  33  mnx^ 

49. 

\a    50.      b 
3a           lb 

51.     f  c  52. 

Id 
Id 

53.  ^h 

54.       %k 

55. 

^m   56.  ^ 

n   57.  .2  a    58. 
n          .3a 

.04  6 
.05  6 

59.       5.67 
-2.31 

X   60.      2.0011/ 
X          —.102?/ 

61. 
62. 
63. 
64. 

0.7  c  +  0.02  c. 
SAd-\-  0.05 ->? 
1.1  g  +  .11^. 
10.1a;-  1.01a;. 

65. 
66. 
67. 

68. 

.682^  + 0.25  y. 
1.001  z  +  9.099 
100.1  w—  1.001 
11.01a—  10  11 

w. 

a. 

76  FIRST  COURSE   IN   ALGEBRA 

Add  the  following,  with  reference  to  the  similar  parts. 

In  case  two  unlike  terms  contain  the  same  letter  or  factor,  we  may 
perfoi-m  the  addition  loith  reference  to  this  letter  as  a  summand,  re- 
garding the  remaining  factors  as  coefficients. 

Ex.  69.   Find  the  sum  of  xz  and  i/z. 

Regarding  x  and  y  as  coefficients  of  z,  the  sum  may  he  expressed  by 
writing  the  sum  of  x  and  y  as  a.  coefficient  of  z,  as  follows : 
xz-^yz  =  (x  -\-  y)z,  or  in  vertical  arrangement,  xz 


(x  +  y)z. 

70.  ax 

71. 

CIJ 

72. 

mz 

73. 

Sw 

bx 

dy 

2z 

yw 

74.  ad 

75. 

6A 

76 

ex 

77. 

3^ 

d 

bh 

-dx 

-hk 

78.  2ab 

79. 

aH 

80. 

x^w 

81. 

X 

36 

bH 

—  y'^w 

^ 

82.  be 

83. 

ab 

84. 

xy 

85. 

&cd 

b_ 

be 

yz_ 

hbd 

86.  2  ax 

87. 

5 

xy 

88. 

-dab 

89. 

12  abc 

Sbx 

-2 

xz 

—  4ac 

—  5  abd 

Compressions  which  contain  a  common  binomial  or  a  commcm  poly- 
nomial factor  may  be  added  with  reference  to  this  common  factor. 

We  may  regard  thefactm^s  which  are  not  common  as  being  coeffi- 
cients with  reference  to  the  common  factors  and,  finding  their  sum  as 
positive  and  negative  numbers,  prefix  this  as  a  coefficient  to  the  com- 
mon facto?'. 

Ex.  90.   Find  the  sum  of  a(x  -\-  y)  and  b(x  +  y). 

Regarding  a  and  h  as  coefficients  of  the  common  factors  of  the  two  given 
terms,  we  may  write  their  sum  as  a  coefficient  to  the  common  part. 
<x  +  y)-\-  h(x  +  y)  =  (a  +  ft)  (rr  +  y). 

Ex.  91.   Ymdthesumof3{x^  —  y^),4:a(x^-y^),a,nd-2b(x^-y^. 
Adding  the  coefficients  3,  4  a,  and  —  2  6  with  respect  to  the  common 
factor  x^  —  y^,  we  have 

3(a;2  -  2/2)  +  4  a(x^  -  y^)  -  2  b(x^  -  t/^)  =  (3  +  4  a  -  2  6)  (x^  -  7/2). 


ADDITION  AND   SUBTRACTION  77 

92.  «:(c  -  d)      94.     x{y  -  z)        96.  2  ab{g  -  k)       98.  a(ir^  -  y^ 
y{c  -  d)  -iy-z)  3  cdjg  -  ;^)  K^^-^/") 

93.  a{b  +  c)       95.  2  a(w  +  w)      97.     a\x  +  3/)        99.  —  (c^  +  1) 

(6  +  c)  bjm  +  n)  -  h\x  +  y)  2d{c^  +  1) 

Subtraction  of  Monomials 

27.  The  algebraic  difference  of  two  numbers  or  quantities  may 
always  be  indicated  by  writing  the  subtrahend  after  the  minuend, 
separating  the  two  by  the  sign  of  operation  for  subtraction  — ,  each 
being  preceded  by  its  proper  quality  sign. 

From  the  principles  affecting  operations  with  positive  and  nega- 
tive numbers  it  appears  that  instead  of  a  subtraction  of  a  number 
we  may  substitute  the  addition  of  a  number  equal  to  it  in  absolute 
valvs  but  of  opposite  qu^ility. 

That  is,  to  subtract  a  given  number  or  term  we  change  its  sign  of 
quality  and  then  proceed  as  in  addition. 

As  in  the  case  of  addition,  this  collecting  of  several  numbers  or 
quantities  into  one  algebraic  expression  is  what  in  algebra  is  called 
subtraction.      The  resulting  expression  is  called  the  difference. 

Ex.  1 .    From  5  x  subtract  2  x. 

Indicating  the  process  by  the  horizontal  arrangement,  we  have 
5  a:  -  2  X  =  (5  -  2)a:  =  3  x. 

bx 
When  employing  the  vertical  arrangement^  2x,  instead  of  actually  altering 
the  sign  of  quality  of  the  subtrahend  2  x,  we  should  make  the  chan<]je  men- 
tally when  performing  the  operation,  and  write  the  result  immediately 
underneath  as  in  arithmetic. 
Ex.  2.   From  —  Sy  subtract  7 y. 
Using  the  horizontal  arrangement,  we  have 

-^y-(-h7y)=-Sy-7y=(-3-7)y  =  -lOy. 

-3y 
Using  the  vertical  arrangement,  we  have  +  7  y 

28.  Caution.  The  student  should  be  careful,  when  employing 
the  vertical  arrangement  for  subtraction,  not  to  actually  change  the 
sign  of  the  subtrahend  on  paper.  Such  a  change  of  sign  is  con- 
fusing when  the  work  is  reviewed. 


i 


78  FIRST  COURSE  IN   ALGEBRA 

29.  It  is  customary,  whenever  possible,  to  so  arrange  an  expres- 
sion as  to  have  the  first  term  preceded  by  a  positive  rather  than  a 
negative  sign. 

E.  g.  We  should  consider  —  6  +  a  to  be  arranged  in  better  order  if 
written  +  a  —  6. 

Mental  Exercise.  VI.    2 


Perform  the  following  indicated  subtractions  : 

1. 

6a 

13. 

-15z 

25. 

2Aa^b 

37. 

45  a^bc^ 

4a 

z 

—  Ua% 

54  a'bc^ 

2. 

96 

14. 

—  \lw 

26. 

-Slcd^ 

38. 

-  56  xSfz 

66 

\lw 

-Ucd' 

67  xYz 

3. 

lie 

15. 

-  18A 

27. 

42  m^'n^ 

39. 

—101  a^xy 

3c 

-\2d 

16. 

—  18A 
2a» 

28. 

-31wV 

40. 

—  Iha^xy 

4. 

-58^y 

123  m^niv 

-    4d 

a« 

46  .ry 

95  mhiw 

5. 

-  \0e 

17. 

5  6« 

29. 

16«6c 

41. 

1 58  S1/W 

—    le 

7  6« 

-  ISabc 

85  syw 

6. 

-  137W 

18. 

4<f» 

30. 

—  25  cdij 

42. 

—IdSgnq 

5m 

-\\d^ 

llcdij 

89gnq 

7. 

Un 

19. 

—  19.r' 

31. 

5Gbgm 

43. 

—150  chw 

-  8w 

—  16  5^ 

20. 

—  23  x^ 

32. 

—  1 7  bgm 

44. 

dlchw 

8. 

21?/* 

-239  6F 

-12^ 

-  19. ?y* 

2^a''bc 

167  6F 

9. 

5r 

21. 

6«6 

33. 

e2ab^G 

45. 

2a 

7r 

27  a6 

—47  aPc 

ia 

10. 

85 

22. 

31  ac 

34. 

-llabc^ 

46. 

5b 

105 

23. 

—  26  «c 

35. 

-38  abc" 
76  a^h 

47. 

hb 

11. 

-  29  6^ 

\c 

-8^ 

^\hd 

—  67  a'^b'^c 

i^ 

12. 

-4y 

24. 

S3  mw 

36. 

-  39  ahh^ 

48. 

f^" 

6./ 

—  27  miv 

WSab'^c^ 

\d 

ADDITION  AND   SUBTRACTION  79 


49.  ^k 

53. 

.hx 

57. 

l.Ga 

61. 

.36^ 

\h 

,2x 

.la 

3.64  ^r 

50.       ^k 

54. 

'-iy 

58. 

2.28  6 

62. 

.345^ 

-\k 

.^y 

.02  6 

-.655^2? 

51.  fM 

55. 

Mz 

59. 

.33  c 

63. 

-    .583  7/ 

1  w 

.04.  z 

.03  c 

-  5.417  y 

52.-^71 

56. 

.Ww 

60. 

2.51^ 

64. 

1.001  z 

^n 

.03  w; 

.25^ 

-.011  2; 

In  case  two  unlike  terms  contain  the  same  letter  or  factor,  we 
may  perform  the  subtraction  with  reference  to  this  letter  or  factor, 
regarding  the  remaining  factors  as  coefficients. 

Ex.  65.     From  ax  subtract  hx. 

Since  a  and  h  are  the  coefficients  of  a:,  we  may  express  the  difference  by 
writing  the  algebraic  difference  (a  —  h)  as  a  coefficient  of  the  common  letter 
a:,  or  {n  —  h)  x. 

Perform  the  following  subtractions  with  reference  to 
similar  parts: 


\ 


66. 

ac 

71. 

hw 

76. 

--  bnq 

81.  -7^/ 

be 
dy 

72. 

—  nw 
bm' 

77. 

—  hnq 

/ 

67. 

baxyz 

82.  ab 

xy 

hm^ 

2  bxyz 

be 

68. 

aw 

73. 

-  9f 

78. 

2ab 

83.  xy 

mw 

—  my^ 

b 

xz 

69. 

-  ct 

74. 

axy 

79. 

-Scd 

84.      xY 

-dt 

bxy 

-d 

-fz' 

70, 

-ay 

75. 

rn^xw 

80. 

hx'^y 

85.  xyz 

xy 

—  n^xw 

-y 

yzw 

The  difference  between  two  compound  expressions  which  con- 
tain the  same  polynomial  factor  may  be  found  with  reference  to 
this  polynomial  factor. 


80  FIRST  COURSE   IN  ALGEBRA 

Ex.  86.     From  a  (x  +  y)  subtract  —h  {x  +  y\ 

Regarding  a  and  —  6  as  coefficients  of  the  binomial  factor  {x  + 1/)  we 
may  find  the  difference  by  prefixing  the  algebraic'difference  of  a  and  —  6, 
as  positive  and  negative  numbers,  as  a  coefficient  to  the  common  factor  (x  +  y) 

as  follows : 

a  (x  +  y) 

-^(a^  +  y) 
(a  -f  6)  (X  +  y) 

87.  5  (a  -f  ^>)  93.       h{m  —  x)  99.  xij(z  —  w) 
2(a  +  b)                       --c{m  —  x)  z{z  —  w) 

88.  8  (c  —  a)  94.      a{c  +  1)  100.  ab{a  —  be) 
—3  (c  —  g)                    —b(c  +  1)  c(a  —  be) 

89.  —6  (tw  —  n)  95.  a(6  +  1)  101.    x(xy  —  z) 
d  (m  —  n)                    (b  4-  1)  yzjxy  —  z) 

90.  -  18  0  +  ^)  96.  m{n  —  1)  102.      ax(bx  -  cij) 
-lS(g  +  k)                -(n  -  1)  -  byjbx  -  cy) 

91.  x{a  +  &) 
y(«  +  b) 

92.  c(7w  —  w) 
d{m  —  w) 

30.  A  polynomial  is  said  to  be  reduced  when  its  like  or  similar 
terms  have  all  been  combined,  as  fiir  as  possible ;  that  is,  when  it 
contains  no  similar  terms. 

Ex.  1.     Reduce  3a:  +  2?/  +  52  —  2a:-3?/  +  7!2  +  4?/to  simplest  form. 

We  have        Zx  +  ^y  +  bz -^x -^y  +  1  z  +  4y  =  x  ■\-'^y  -\-l2z. 

Exercise  VI.     3 
Reduce  each  of  the  following  polynomials  to  simplest  form: 

1.  2a  —  46  +  6c  —  a  +  56  —  2c. 

2.  bx-\-ly  —  2z  —  4:X  —  ^y  +  z. 

3.  lla  —  2d-\-?,b-\-^d-b  +  a. 

4.  12  a  — 6  +  ^+3c  +  2'6  — 2c  — a  +  3  6. 


94. 

a{c  +  1) 
-b{e  +  1) 

95. 

96. 

a(b  +  1) 

m(n  -  1) 

-(7^-1) 

97. 

(^-1) 

x(x  -  1) 

98. 

2(c  +  g) 
3b(c+g) 

ADDITION   AND  SUBTKACTION  81 

5.  13.r—  13?/+  140—  U2V  —  2Z  +  2  i/ —  2w  +  4:Z. 

6.  6a:  +  i/-'l0z  —  Si/  —  2ic+  dz-\-  4:X—  5i/  +  8z, 

7.  Sa  +  5b  —  c+2  —  2a  —  4:b+ec+7  —  a. 

8.  5k  —  4:7n  —  8n  +  ^  —  Sk  +  4:m  +  8n  —  d  —  2k. 

9.  4.a^-2ab+  10 c^  +  S ab -\- a'' -{-  2b''-3c^-ab  +  Qb"". 
10.  9  ^^  -  ^y  +  3  -  5^'  +  8/  +  ^^  +  5  -  4^2  -  8f. 

31.  The  Check  of  Arbitrary  Values.  Since,  unless  the  con- 
trary is  expressly  stated,  the  result  of  any  operation  with  letters  is 
obtained  without  restricting  the  values  of  the  letters,  we  may  as- 
sume that  they  have  such  values  as  we  choose  to  assign  to  them. 
This  simple  check  of  arbitrary  values  will  be  found  in  most  cases 
to  be  sufficient  to  indicate  errors  in  a  calculation  if  there  be  any. 

Whenever  numerical  checks  are  used,  it  should  be  understood 
that  the  substitutions  are  to  be  made  in  the  example  as  originally 
given,  and  also  in  the  final  result.  All  intermediate  steps  leading 
from  the  first  indicated  operations  to  the  final  result  should  be 
neglected.  If  the  original  indicated  operations,  when  performed 
with  numerical  values,  produce  the  same  result  as  is  found  by  sub- 
stituting numerical  values  in  the  final  result  —  that  is,  if  oar 
work  "  balances  "  —  we  have  a  check  (except  in  certain  very 
special  cases)  upon  the  accuracy  of  all  of  the  intermediate  steps 
which  have  been  neglected  in  the  checking  process. 

If,  in  checking  examples,  the  value  1  be  assigned  to  any  particu- 
lar letter,  errors  among  exponents  may  not  be  detected,  since  all 
integral  powers  of  unity  are  1. 

32.  A  clieck  upon  an  operation  is  another  operation  which  is 
such  as  to  verify  the  result  first  obtained. 

E.  g.  Such  a  check  is  the  employment  of  addition,  to  verify  an  example 
in  subtraction,  or  the  multiplication  of  a  divisor  by  the  quotient  to  obtain 
the  dividend  in  an  example  in  division. 

Addition  of  Polynomials 

33.  To  add  a  polynomial,  it  is  sufficient  to  add  its  terms  succes- 
sively. 

When  adding  two  polynomials  it  will  be  found  convenient  ./?r5^  to 
(ir range  tJie  terms  according  to  the  jjowers  of  some  letter  of  reference^ 

G 


82  FIRST  COURSE  IN   ALGEBRA 

and  then  to  write  the  polynomials^  one  under  the  other ^  in  such  a  way 
that  similar  terms  shall  be  in  the  same  vei'tical  column. 

Then  add  the  columns  separately  and  connect  their  sums  by  the 
resulting  signs. 

Ex.  1.   Add  3a:y«  —  3  x'^y  +  a;«  —  y^,  x^  +  2  ?/«  +  2  x'^y,  and  —  a;^?/  +  4  xy^ 

Arrange  the  polynomials  according  to  descending  powers  of  x,  and  write 
them  so  that  similar  terms  shall  appear  in  vertical  columns,  as  below. 
Adding  the  first  column,  we  have  2  a:*  ;  adding  the  second  column,  —2x^y  ; 
adding  the  third  column,  7a;^^;  in  the  fourth  column  the  sum  is  0. 

Hence  the  resulting  sum  is  2  x*  —  2  x'^y  +  7  xy"^. 

Check.     Let  X  =  2,  y  =  3. 

x8-3x2y +  3xy2-     y^ -1 

a:«  +  2x'^  +2^8 §6 

-    x'^y-if^xy^-    y8 ^ 

118 

2x»-2x2y  +  7xya  118 

~~0 

We  may  check  the  result  by  substituting  the  values  2  and  3  for  x  and  m 
respectively  in  the  given  expressions,  obtaining  the  values  —  1,  86  and  33, 
as  shown  above. 

The  algebraic  sum  of  these  values,  118,  should  "  balance  "  with  the  result 
obtixined  by  substituting  the  same  values,  x  =  2  and  2/  =  3,  in  the  result  of 
the  algebraic  addition,  2x'*  —  2x^y  +  1  xy^.  This  value  is  found  to  be  118, 
and  according  to  this  test  the  work  is  correct. 

34.*  Detached  Coefficients.  In  performing  an  example  in 
addition,  when  writing  several  similar  terms  in  a  column,  it  is  not 
strictly  necessary  to  write  the  literal  factors  every  time,  provided 
that  it  is  understood  that  they  are  the  same  as  those  of  the  term  at 
the  head  of  the  column. 


E.  g.   Instead  of        +5  a%\    write    +  5 
-  3  a%'ic  -  3 

+  2  amc  +  2 


a%^c 


Aa%^c  4    a%^c 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


ADDITION   AND  SUBTRACTION 


83 


Ex.  2.   Add  the  following,  using  Detached  Coefficients  : 
dx^+^xy  +  bxj/^—y'^,  3xy —  2xy'^  +  x^—2y%  iind2x^+2x"  —  xy +  3xy'^. 

Check.    Let  x  =  2,  y  =  4. 


+  2 


x'^  +  2 
+  3 
-  1 


xy  +  5 
-2 
+  3 


Xy2_ 


-2 


4   a:8  +  5   x"^ -\- 4   xy  +  Q   xy^-  3   y"^ 


+124 

-160 

+120 

84 

84 

0 


Exercise  VI.    4 

Perform  the  following  indicated  additions;  the  answers  to  the 
first  twenty-eight  examples  may  be  obtained  mentally  : 


V 


1.  3a  +  26 
a  +  56 


7.  ^m—Wn 
4:m  +  14  w 


13.  -  lOA—  15y 
—    3  ^  +    4  7/ 


2. 

6a- 

96 

8a- 

-36 

3. 

76- 

He 

46- 

19c 

4.  Qd+       g 
6cg—  12gr 

5.  5a  +  26 
3a  4-     6 


6. 
19. 

6  c  — 3</             12 
4c-2c? 

8a  +  66  — 4c 
-3a-56+  7c 

20. 

2a+    56-    7c 
3  a—  116  +  14  c 

21. 

46-  12c  — 24d 
66+  12c—    7  c? 

8. 

Ux+  21y 
32^-23y 

9. 
10. 

7a  +  6 
2a  +  6 

96  — 3c 
96  +  8C 

11. 

6x-ly 
—  5x  +  ly 

12. 

3a-Qb 
3  a  -  6  6 

14. 

—  Urn  —  20q 

—  llm  —  llq 

15. 

15  a6  —       cd 
-23a6  +  2^cd 

16. 

0  a— 14  be 

—  26  a  +  33  6c 

17. 

2  a  +  3  6  —  4  c 

a  —  5  6  +  7  c 

18. 

X  —  Sy  +  5z 
—  2x-\-4y  —  &z 

22.  9x  —  Sy+3z 
3x  —  Sy+  9z 

23.  10a  -  76  +  12c 

3a—  76  -  10c 

24.  3^-4?/+  5z 
6x  +  4i/  —  Sz 


84  FIRST  COURSE   IN   ALGEBRA 

25.  4a«+ 9^*2- 16  27.  13a' -  17  ^^'^  +       c" 

ba^+^b^-    1  41«'^-21/>'-33c^ 

26.10  a—       m-\-lbw  28.  11  ar' +  13;r3/ +  21/ 

-  a+  19m-28  2g  '  11  ^r'^  -  31  a-y  -  12/ 

Find  the  sums  of  the  following  groups  of  expressions  : 

29.  a  +  26+c;3a  +  6  +  c;a  +  ^  +  4c. 

30.  2x -\- y -\-^z)  x-\- 4:y-\- z;  bx-\-Qty-\-Sz. 

31.  56  — c  + 2c?;  36  — 2c+ 4c?;  6  — 3C  +  6?. 

32.  ^X'-^y  —  2z)lx—\0y  —  ^z]4tx—by  —  %z. 

33.  7a  —  2  6  +  6a;;5a  +  86  —  4;r;a  —  96  —  3;r. 

34.  56  + 9c?-4m;;  86  — 7c?— 3?^;  66  — 2c?-2^. 

35.  g  —  ^h-\-^k\lg  —  h  —  2k\'6g-^bh  —  (Sk. 

36.  7»  —  3w  +  7y;  5w  —  8w  +  2y;  97W  —  4w  +  6y. 

37.  2a  — 56  — 3c;  —9a  — 6  — 4c;  8a  +  66  + 7c. 

38.  6;r  +  3y  — 22;;  — 4;r— 7i/  — 82;;  — ^+ 5^^+ 92?. 

39.  3a  —  2  6  +  4C—  10;36  —  2c  +  8;— 2a  —  c+1. 

40.  2/  -  c'  +  10 ;  5=*  —  w'  —  9  ;  /  +  w;^ 

41.  47W«  +  ;r^- 7;  —  67W»  +  4a?;  7w«  — ^  — 3;r+ 7. 

42.  c«-4cV-2cc?2  + 2c?«;  3c^c?+ 2c(^;  c'(/-^«. 

43.  9/^  -  6m'  -  7 joV»  -  (/^  -  8 joV  +  6 J0(7'  +  77>V. 

44.  4  a6  —  2  ac  +  10  a6c ;  —  6  a6  —  9  a6c  +  2  6c ;  3  a6  +  4  ac  — 
a6c  —  3  6c. 

45.  7;r«  +  8;r— 11;  — 4;r'+12;  —ha^—^x  —  2;  —  x""  + 
hsi^-x^-% 

Subtraction  of  Polynomials 

35.  To  subtract  a  polynomial  we  may  subtract  successively  the  terms 
of  which  it  is  composed.  Hence  we  may  reverse  the  sign  of  quality  of 
each  of  its  terms  and  then  proceed  as  in  addition. 

Ex.  1.         Subtract  Ax^  +  Qxy  -ly^  from  Qx^  +  xy  -2  y^. 

We  may  indicate  the  operation  by  writing 

6x2  +  a;y  -  3/ -  (4 a:2  +  9a;y  -  7 2/2). 

Removing  the  terms  from  the  parentheses  preceded  by  the  minus  sign, 
and  reversing  the  signs  of  quality  of  the  terms  removed,  we  have 
6x2 +  a:y-32/2_  (42-2  _|.9a.^_7y2)  =.Qx^-\-xy  -Zy'^  -  4:x'^ -9xy +  1  y"^ 

Combining  like  terras,  =  2  x2  —  8  x^  +  4  5/2. 


ADDITION  AND  SUBTRACTION  85 

Whenever,  as  in  the  example  above,  the  operation  of  subtraction 
is  indicated  by  enclosing  the  subtrahend  in  parentheses  preceded  by 
a  minus  sign,  we  may  remove  the  parentheses  and  actually  change 
the  signs  of  the  numbers  removed. 

If,  however,  the  subtrahend  is  simply  written  underneath  the 
minuend  and  we  are  directed  to  subtract  one  expression  from  the 
other,  it  is  better  not  to  change  the  signs  on  paper,  but  to  make 
the  change  mentally  while  performing  the  calculation.  (Compare 
with  §  28.) 

The  following  arrangement  is  usuaUy  more  convenient : 

Check.     Let  a:  =  4,  y  =  3. 
6a;2  4-     a:y-3y2 92 

4x2  +  9x^-7  3/2 110 

^18 

2x2-8x3/ +  4^2 _18 

~0~ 

Exercise  VI.    5 

Perform  the  following  indicated  subtractions,  reversing  the  signs 
mentally ;  the  answers  to  the  first  sixteen  examples  may  be  obtained 
mentally  : 


1. 

2a  +  b 

7. 

-2m  — ^ 

13. 

—  ax  +     by 

a-b 

—  37W  — 4 

2  ax  — 2  by 

2. 

3a  +  46 

8. 

2  +     a 

14. 

3a  +     6  +  c 

2a-56 

3  — 2a 

2a  +  2b-c 

3. 

4.r-  ly 

9. 

—  l()x  —  y 

15. 

—  4:X-5y+Qz 

bx—  6y 

—  llx  +  y 

5x  +  Gy+lz 

4. 

76+     c 

10. 

—  ac—    h 

16. 

8a-  76  +  6C 

76  — 2c 

2ac  +  2b 

-6a+  76  — 8c 

5. 

6r  —  5 

11. 

4:X  —  by 
—  Sx  —  Gy 

6. 

2a+     a: 
—  ^a  —  2x 

12. 

xy  —  zw 

—  xy  +  zw 

86  FIRST  COURSE   IN  ALGEBRA 

17.  From  6  a  —  10  ^  —  7  subtract  4  a  —  5  6  —  2. 

18.  From  5a  +  Sb  —  d  subtract  2  a  +  3  6  —  4. 

19.  Subtract  a^+2ab  +  0^  from  2  «"  -  2  <76  +  2  b\ 

20.  From  Sar^  +  4:xy  +  8  subtract  4^^  —  "Ixy  —  3. 

21.  Subtract  x"  -^^xy^-xf  from  4  ^r^  +  10  xy  +  13  j/^. 

22.  Subtract  x^  -\-  a^y -\-  xf  +  y*  from  x^  +  ?/^ 

23.  From  a»  +  3  a"^  +  3  al)"  +  ^*  subtract  «»  +  V. 

24.  Subtract  w*  —  3  m^n  +  3  tw/^^  —  n^  from  w^  —  7j*. 

25.  From  a*  +  rt^  +  a  subtract  —  2  «*  —  3  f<^  +  a. 

26.  From  ah  ■\-  he  ->r  ca  subtract  —  ah  -^  he  —  ca. 

27.  Subtract  2  a*^  —  ah  —  5h^  from  3  ^^  +  ^^  —  4  ^^. 

28.  Subtract  6^^  -  1  xy  —  8/  from  Ix'-^xy—lf. 

29.  Subtract  5  xY  -  10  ;r/  +  16  ^  from  5  ^y  +  10  ^/  +  1 6  ?/. 

30.  Subtract  a''  +  3  a^^  +  3  ah^  +  6»  from  a«  -  3  a^'h  +  3  aU^  -  h\ 

31.  From  x^  ■{■  b  x^y  -{■  &  ^y  +  3  ;r/  +  /  subtract 

5^V+  6^/  +  3;r/. 

32.  Subtract  c*  -  c^^  +  c?«  from  3  c»  -  c'c?  +  4  cc?^  +  5  c?«. 

33.  From  5  ^^  +  6  ^"^^  +  3  ^'^  +  7  A^  subtract  3  ^^  +  6  /A'  -  5  h\ 

34.  Subtract  4  a*  +  7  a''^  +  6  a^'b''  +  6*  from 

4a*+  lOa'6-  15 a'^^^^-  9  6*. 

35.  From  12  a»  +  190^^  +  17a&^  +  21  6*  subtract 

5a«+  lla'^h-ah''-  20h\ 

36.  From  13  d^  —  23  fi?V  +  11  c^r^  +  14  r^  subtract 

13d^-24(^V+  10c?V+  14r^. 

Wben  we  are  required  to  subtract  tbe  sum  of  several  polynomials 
from  the  sum  of  several  others,  we  may  treat  the  problem  as  one  of 
addition  by  actually  changing  the  signs  of  all  those  expressions 
which  are  to  be  subtracted  and  then  proceeding  with  the  resulting 
expressions  as  an  example  in  addition. 

37.  From  the  sum  of«  +  26  +  3c  and  3 <2  —  &  +  4 c,  subtract 

2a  —  h  —  6c. 

38.  From  thesumof  5^  — ?/  +  42;and  4^  — 2?/  — 3^;,  subtract 

3^  +  ^y  —  bz. 

39.  Subtract  3a  +  2h  +  26  —  3(1  from  the  sum  of 

a  —  b  +  c  —  d  and  2a  —  b  +  2c  —  d. 


i 


l< 


ADDITION   AND  SUBTRACTION  87 

40.  Subtract  7a  —  3b  —  c  —  d  from  the  sum  of 

Qa  —  lj  +  4:C  +  d  and  Sa  +  4:b  —  2c  +  d. 

41.  From  the  sum  of  a^  +  ab  +  b^  and  3a^  —  2ab  —  4rb%  subtract 

2a^  —  Sab  —  Sb\ 

42.  Subtract  ab  +  Sbc  —  4:cd  from  the  sum  of  Qab  —  4:bc  +  cd 

and  —  4:ab  +  2bc  —  Scd. 

43.  From  the  sum  of  ab  +  be,  be  —  cd  and  da  —  cd,  subtract 

ab  +  be  —  cd  +  da. 

44.  From  the  sum  of  3 aft  +  bcy  —  ^bc  -\-  cd  and  2da  —  2  cd^ 

subtract  2ab  —  2bc  -^  cd  -\-  da. 


88  FIRST  COURSP]  IN  ALGEBRA 


CHAPTER  VII 

MULTIPLICATION  OF  INTEGRAL  ALGEBRAIC  EXPRESSIONS 

(For  Definitions,  See  Chapter  V) 

Principles  Relating  to  Powers 

1.   We  have  already  defined  a"  to  mean  the  product  of  n  factors 
each  equal  to  a.     Accordingly,  we  have  as  a 


-n  f actora- 


(i.)   Defiuition  Formula:  a"  =  aX«XaX Xa, 

n  being  understood  to  be  a  positive  whole  number,  and  a  being 
different  from  zero.     (See  Chapter  VI.  §  5). 

E.g.   26  =  2x2x2x2x2;    6^  =  ay.  ay.  a. 

Product  of  Powers  of  the  Same  Base 

As  a  direct  consequence  of  the  definition  formula  above,  we  have 
for  equal  bases  the  following 

(ii.)   Distribution  Formula  a"*  X  cC  =  a"'+". 

If  a  be  any  number  other  than  zero,  we  may  indicate  the  product 
obtained  by  multiplying  «"*  by  «"  by  writing 

/ m  factors > n  factors n 

dT"  y.  d!"^  {aaa a){aaa a)  =  «"*+". 

The  principle  expressed  by  the  distribution  formula  may  be  stated 
as  follows  : 

Law  of  Indices.  The  product  obtained  by  multiplying  a  power 
of  a  given  base  by  another  power  of  the  same  base  is  a  power  of  the 
same  base,  the  exponent  of  which  is  found  by  adding  the  exponents  of 
the  factors. 

E.  g.   h^xb^  =  65. 

In  order  to  find  the  power  of  a  power  of  a  base  we  may  employ 
the 


MULTIPLICATION 


-n  factors- 


(iii.)   Association  Formula  (a^'Y  =  «"*  X  «"*  X  «"'  X X  a" 

/— n  terms , 


TTiat  is,  the  power  of  a  power  of  a  base  is  a  power  of  the  base,  the 
exponent  of  which  is  found  by  multiph/ing  the  exponent  of  the  given 
pov)er  of  the  base  by  the  exponent  of  the  required  power, 

E.  g.  (c3)2  =  c«. 

Powers  of  Products  of  Different  Bases 
(iv.)   Distribution  Formula     (ahy  =  oTir. 

From  the  definition  of  a  power,  we  have 

f TO  factors — \ 


{abY  =  (ab)  X  (ab)  X  (ab)  X X  (ab), 

f TO  factors s         / -«i  factors- 


=  aXaXaX X  aXbxb  Xb  X X  b, 

=  drv^. 

The  w'*  power  of  a  product  of  several  faxitors  may  be  written  as  the 
product  of  the  n*^  powers  of  these  factors,  and  conversely. 

Ex.  I.         (2  ay  =  2hi^  Ex.  3.  (-  c)8  =  (-  lyc^ 

=  8  a'.  =  -c». 

Ex.  2.  (5  64)2  =  52(j4)2  Ex.  4.  (-bx)^=  (-  5yx^ 
=  25  68.  =  25  xK 

It  may  be  shown  that  the  laws  above  hold  for  more  than  two 
exponents  or  for  more  than  two  factors.     That  is, 
Corresponding  to  (ii.),  above,  we  have 

a"*  X  a"  X  aP  X X  a""  =  a'"+"+''+ +«'. 

Corresponding  to  (iii.) 

(((aryyy =  a^^"*"^ 

Corresponding  to  (iv.) 

(abed »)'"  =  a"*  X  6"*  X  c"*  X  ci"*  X Xaf. 

Mental  Exercise  VII.     1 

Express  each  of  the  following  products  of  different  powers  of  the 
same  base  as  a  single  power  of  the  given  base  : 


90  FIRST  COURSE   IN  ALGEBRA 


1.  2*  X  2^ 

12.  /^«  X  k\ 

23.  w^  X  ??«  X  w^ 

2.  4'  X  4. 

13.  >t^  X  k\ 

24.  ;^'"  X  ^"^  X  .^^^'». 

3.  3*^  X  3. 

14.  w'  X  w^^ 

25.  a^'"  X  ««'". 

4.  5*  X  5. 

15.  w«  X  w». 

26.  b"'  X  b^\ 

5.  2«  X  2\ 

16.  ;r«  X  .r'. 

27.  c^'"  X  (f. 

6.  a^  X  a*. 

17.  a^  X  a«  X  a*. 

28.  c?"-=^  X  ^. 

7.  ^»  X  b\ 

18.  b"^  X  b^  X  b\ 

29.  A"+^  X  A"-^ 

8.  c*  X  c\ 

19.  c*  X  c*  X  c'. 

30.  af'^'^^  X  af-'-K 

d.  d^X  (P. 

20.  (P  X(f'  X  d*. 

31.  ^  X  ^  X  ^. 

10.   e^  X  e\ 

21.  ^«  X  ^-^  X  k. 

32.  /"  X  7/8"  X  7/^^ 

11.  5^'  X  5r». 

22.  w^  X  w^  X  m^^. 

33.  ;5^'-  X  s''  X  z^'. 

Express  each  of  the  following  powers  of  ] 

powers  as  a  single  power 

of  the  given  base  : 

34.  (2y. 

44.  (g«)*. 

54.  (^y\ 

35.  (sy. 

45.  (gy. 

55.  (ay. 

36.  (5^. 

46.  (^*)^ 

56.  (6»)^. 

37.  (6'*)^. 

47.  (A^)«. 

57.  (c")». 

38.  (10^». 

48.  (ky. 

58.  (dy. 

39.  (7^'. 

49.  (m'^)^ 

59.  (A'")'". 

40.  (a^. 

50.  (ny. 

60.  (F)». 

41.  (by. 

51.  (;.«)«. 

61.  (w'^^)^ 

42.  (c^y. 

52.  (/)'. 

62.  (w«'')«>'. 

43.  (d^y. 

53.  (f^y\ 

63.  (x^y. 

Express  the  following  powers  of  products 

as  products  of  powers  : 

64.  (Say. 

76.  —  (5^«^0*- 

88.  -(la^ifzy. 

65.  (4.  by. 

77.  -(-Sacy. 

89.  -(-4^*)*. 

66.  (Qcy. 

78.  (a'^^)'. 

90.  (aby. 

67.  -  (2  d)\ 

79.  (bey. 

91.  (/.c)'^. 

68.  (3  my. 

80.  (c^^)^ 

92.  (a'^by. 

69.  (-  7  ^)^. 

81.  (w"w)«. 

93.  (c^^^)*-. 

70.  (-  4:i/y. 

82.  (a^yy. 

94.  -  (xyy. 

71.  (-2  4'. 

83.  (a^/^c)^ 

95.  («"^^)'". 

72.  -(2aby. 

84.  -  (a%cy. 

96.  (afyy. 

73.  (3  6c)^ 

85.  («^'^c«)^ 

97.  -  (a^^^'')'^. 

74.   (2xyy. 

86.  (a'b'cy. 

98.   (a'^'lfy. 

75.  (-3?/^)^ 

87.  (-2ciVV)l 

99.  (^^«0    , 

MULTIPLICATION  91 

Express  the  following  products  of  powers  as  powers  of  products  : 

100.  2»  X  3^  105.  h^  X  c\  110.  2bxYz\ 

101.  5'  X  2^.  106.  w*  X  n\  111.  32 a^b^a^. 

102.  3'  X  7^.  107.  a^  X  b^  X  c\  112.  64  aV;?^ 

103.  e  X  3^.  108.  ^  X  /  X  ;:'.  113.  (-  ayb\ 

104.  a""  X  b\  109.  Sa^b^.  114.  [- x)Y^. 

2.  Product  of  Two  or  More  Monomials.  The  product 
obtained  by  multiplying  one  monomial  by  another  is  the  monomial 
obtained  by  multiplying  together  all  the  factors  of  the  two  in  any 
order. 

Hence,  to  multiply  one  monomial  hy  another,  determine  first 
the  sign  of  tJie  quality  of  the  product.  TJien^  for  a  numerical  coeffi- 
cient^ write  the  product  of  the  numeral  factors  of  the  monomials^ 
followed  by  the  ^product  of  the  different  letters^  each  letter  having  for  an 
exponent  the  sum  of  the  exponents  of  this  letter  in  the  two  monomials* 

Proceed  similarly  for  produ^cts  of  three  or  more  monomials. 

Ex.  1.   Multiply  3a;V  by  2a:i/*z. 

Both  terms  are  understood  to  be  positive  since  no  signs  are  written  before 
them.  We  have  as  a  numerical  coefficient  the  product  of  3  and  2 ;  x  occurs 
to  the  2  +  1  or  3rd  power  ;  y  to  the  3  +  4  or  7th  power;  z  to  the  1st  power. 

Hence,  3  xhf  x  2  zyh  =  6  xYz. 

Ex.  2.    Find  the  product  of  5  ab^c  and  —  2  abc^. 

Since  the  terms  are  of  opposite  quality,  the  product  is  a  negative  number. 

Hence,  (5  ab^c)  x  (-  2  abc^^)  =  -  10  a%h^ 

Ex.  3.    Find  the  continued  product  of  3  a%,  2  be  and  —  5  ac. 

Since  two  of  the  given  numerical  coefficients  are  positive  numbers  and 
one  is  a  negative  number,  we  may  write  for  a  numerical  coefficient  the  pro- 
duct of  3,  2,  and  5,  or  30,  prefixing  the  —  sign  to  indicate  its  quality.  The 
literal  factor  a  occurs  2  +  1  or  3  times ;  &,  1  +  1  or  2  times ;  c,  1  +  1  or 
2  times. 

Hence.  (3  a%)  (2  be)  (-  5  ac)  =  -  30  a%^c^. 

The  student  should  check  the  examples  above  numerically. 


92  FIRST  COURSE   IN   ALGEBRA 

Mental  Exercise  VII.   2 
Perform  the  following  indicated  multiplications  : 


1. 

a 

14. 

-he 

27. 

8c^ 

40. 

na^b 

2^ 
3 

15. 

—  a 

28. 

9d^ 

41. 

11  ab^ 

2. 

ah 

-Ix^yz 

b 

cd 

Sd' 

Uyz^ 

3. 

—  c 

16. 

ay 

29. 

10/ 

42. 

l^a'bc 

4 

hx 

-4/ 

-5ahc^ 

4. 

d 

17. 

mw 

30. 

11  A^ 

43. 

IS  xYz 

—  5 

nx 

-Ih' 

Qxyh^ 

5. 

2n 

18. 

a^h 

31. 

13  i^ 

44. 

12«^6V 

3 

c' 

3P 

11  a'h'^c 

6. 

4ar 

19. 

^f 

32. 

1  ah 

45. 

a'^h'' 

6 

4a 

ah 

7. 

-ly 

20. 

i^'y 

33. 

14  6c 

46. 

af^f 

5 

21. 

-n^ 

34. 

5  be 
4  ah 

47. 

xy 

8. 

2a^ 

a^f 

-6 

a 

11  he 

-xy 

9. 

-  llw 

22. 

3h' 

35. 

-6x1/ 

48. 

-a'^'x' 

-8 
c 

23. 

36. 

8xz 

49. 

-  (^1 

10. 

12  mn 

a'%'' 

d^ 
-h 

24. 

-c« 

37. 

12  mx 
15  a^h 

50. 

x'^'h^ 

11. 

-Id^ 

32  a^y'^'^z^'"' 

ah 

25. 

-     <P 
3^^ 

38. 

—  4  he 
11  he' 

51. 

-?>xyz 

12. 

14^2m+l^„-l 

—  c 

—  xz 

26. 

2h 

39; 

9hd 

52. 

-ha^h"- 

13. 

a^-ijf^n+i 

y 

6w 

-  6  m7i^ 

a^+i^V-i 

MULTIPLICATION  93 

53.  ah  X  hex  ca.  60.  xhjz^  X  xifz^  X  y^^w^. 

54.  xyXcczX  xw.  61.  {^a%)(2ac%iohh). 

55.  a^h  X  h^c  X  c^a.  62.  —  (dm^x)(Sm^x)(2mnx). 

56.  ahc  XahXa.  63.  (3  ««6«) (4  ^ V)  (-  5  cV). 

57.  a=^^>'  X  ^V  X  cU^  64.  (7^2;:«)(-  3^/)(5/;^2). 

58.  -  a^x  X  hhj  X  cVy\  65.  -  {<oa^hc)(:2a'c^){^h^&). 

59.  a^bc  X  ab\  X  abc\  66.  (-8xyz)(-  Sxy'')(4:a^z'). 

67.  (-  2  a«^c)(-  4  «ra)(-  3  bc^d^). 

68.  (-5^V)(2^-'^)(-63/-JW^. 

69.  (-  7  a%^j^)(2  a'hyX-  6  b^a^y^ 

70.  (-  9  a%h'){-  5  a%H'){-  2  6V^). 

3.  Two  terms  are  said  to  be  of  the  same  type  when  they  may 
be  derived,  one  from  the  other,  by  interchanging  the  letters  of  one 
or  more  pairs  of  letters. 

E.  g.  The  terms  xhjz,  y^xz  and  z^xy  are  all  of  one  type.  The  second 
expression  may  be  obtained  from  the  first  by  interchanging  the  letters  x  and 
y,  and  the  third  may  be  obtained  from  the  first  by  interchanging  z  and  x. 

Two  expressions  are  said  to  have  the  same  form  or  to  be  of  the 
same  type,  with  reference  to  certain  specified  letters,  if  to  every 
term  in  either  there  corresponds  one  and  only  one  term  of  the  same 
type  in  the  other. 

E.  g.  The  expressions  a'  +  2  a&  +  6^  iind  m^  +  2  mn  +  n^  are  of  the  same 
type. 

4.  When  any  one  term  of  a  particular  type  is  given,  the  literal 
parts  of  all  others  of  the  same  type  for  a  given  set  of  letters  may  be 
written  at  once. 

E.  g.  For  three  letters,  x,  y,  and  z,  the  terms  of  the  type  x^  are  x*,  y^,  and 
2^.  The  terms  of  the  type  xy  are  xy^  yz,  and  zx.  The  terms  of  the  type  xh/ 
are  xh/,  xh,  y^x,  y\  z^x,  and  z'hf. 

5.  It  is  customary  when  writing  integral  expressions  to  arrange 
consecutively  those  terms  which  are  of  the  same  type,  and  to  group 
those  of  the  same  degree.  When  so  arranged  an  expression  is  said 
to  be  in  standard  form. 

E.  g.    The  following  expression  is  in  standard  form  : 

x«  +  ?/»  +  2^  +  x^y  +  xh  +  y^x  +  yh  +  z^x  +  z^y  +  xyz. 


94  FIRST  COURSE  IN   ALGEBRA 

6.  A  variable  term  of  a  polynomial  is  a  term  in  which  one  or 
more  letters  appear.  This  name  is  appropriate  because  the  numerical 
value  of  a  term  changes  or  varies  as  different  numerical  values  are 
given  to  the  letter  or  letters  appearing  in  it. 

If  a  term  appearing  in  a  polynomial  does  not  contain  any  letter, 
but  instead  consists  of  a  numeral,  it  is  called  a  constant  term, 
because  its  value  remains  unaltered  when  numerical  values  are 
assigned  to  the  letters  appearing  in  the  remaining  terms. 

7.  A  polynomial  in  which  the  terms  are  arranged  according  to 
the  powers  of  some  letter  is  said  to  be  complete  when  no  powers  of 
this  letter  are  missing,  from  the  highest  one  contained  down  to  and 
including  the  constant  term. 

E.  g.  The  polynomials  a:«  +  a:^  +  x  -|-  1  and  3  ax*  +  2  ftx*  -  cx^  -\- dx  -  2 
are  both  complete. 

A  polynomial  is  said  to  be  incomplete  with  reference  to  the 
powers  of  a  specified  letter  if,  when  its  terms  are  arranged  according 
to  the  powers  of  this  letter,  one  or  more  powers  are  missing. 

E.  g.  In  x*  —  1  the  powers  missing  are  x^  and  x ;  in  x*  +  x^  +  1  the 
powers  missing  are  x*  and  x  ;  in  x*  +  x'*  +  x  the  constant  term  is  missing. 

Multiplication  of  a  Polynomial  by  a  Monomial 

8.  As  a  direct  application  of  the  Distributive  Law  for  Multi- 
plication in  its  first  form,  we  have 

x(a  4-  6  —  c)  =  xa  +  x&  —  xc.  (1) 

Since  the  value  of  a  product  remains  unchanged  when  we  alter  the 
order  of  the  factors  (Commutative  Law),  we  have 

(a  +  h  —  c)x  =  ax  +  bx  —  ex.  (2) 

From  the  expression  above,  it  appears  that  a  polynomial  Tnay  he 
multiplied  by  a  monomial  by  multiplying  each  term  of  the  polynomial 
by  the  monomial  (observing  the  law  of  signs),  and  taking  the 
algebraic  sum  of  the  partial  products  thus  obtained. 

Ex.  1.    Multiply  3  x«  +  X  -  2  by  3  a6. 

We  may  obtain  the  product  by  multiplying  the  successive  terms  3x2,  ^^ 
and  -  2  by  3  a6  and  finding  the  algebraic  sum  of  the  partial  products  thus 

obtained. 

(3  x«  +  X  -  2)  X  3  a6  =  9  afix^  +  3  a&x  -  6  ah. 


MULTIPLICATION  95 

Instead  of  employing  the  horizontal  arrangement  of  the  work,  as  above, 
we  may  use  the  vertical  arrangement  corresponding  to  the  one  commonly 
used  in  arithmetic,  as  shown  below  : 

Check.     Let  a  =  1,  &  =  2,  a;  =  3. 

Multiplicand  .....  Sx^    +       z  -  2 28 

Multiplier Sab 6 

168 

Product 9abx^+3abx-6ab  ....   168 

0 
9.  It  should  be  observed  that,  since  in  algebra  we  do  not  use  the 
positional  system  of  notation  employed  in  arithmetic,  we  may  write 
the  multiplier  in  any  convenient  position  beneath  the  multiplicand. 
The  partial  products  may  then  be  obtained  by  beginning  either 
with  the  first  term  or  the  last  term  of  the  multiplier  and  multiplying 
the  terms  of  the  multiplicand  successively  by  it,  either  from  left  to 
right  or  from  right  to  left. 

Mental  Exercise  VII.  3 
Perform  the  following  indicated  multiplications  : 

1.  a  +  b                         9.  4a  +  6  17.        x^  —  f 
2 3c —  ipy 

2.  b  —  c  10.  d  —  5h  18.  2  mn+  3  be 
3                                       4  a:  4  a;r 

S.  w  +  y  11.       &m  —  5n  19.  lax +6 b^/ 

z  —2y  Smn 

4..  y  —  z  12.  ab  +  cd  20.        8cd-\-9bh 

w  xy  —  2  kr 

b.  2  m  -\-3n                 IS.  ab  +  ac                  21.  Aax^  +  Sby^ 
4 be 2cd 

e.ek  —  8k  U.xy  —  xz  22.  Sa^b  + 5  ab^ 

5 xy  4  ab 

7.  2x-\-  Sy  15.  ab- be  23.       Qxy^  -  ifz 
—  4 ca  —2  xyz 

8.  7m -n  IQ.  a^  -  b^  24.       4 a.r*  +  3  by^ 
—  8                                 ab  —Sabh 


96  FIRST  COURSE  IN  ALGEBRA 

25.  Qa^z^  +  ha^z  30.  a"  -  2 ah  +  b'' 

4  xyz  3  ab 


26. 

9a^6^c+  12aZ>V 
^abc 

27. 

7.rV 

28. 

a»  +  a^  +  a  +  1 
2a 

29. 

6«  -  ^^  +  6  -  1 
4^^ 

31. 

m^  ^2  mn  +  w* 
—  2mn 

32. 

ab  +  be  •{-  ca 
abc 

33. 

abc  +  abd  +  ico? 
abed 

34.  (a  +  ^  4-  c)  a^c.  39.  (^  —  ^r^  +  ^^^  _  ^s^^  ^^^ 

35.  (j-^  +  r  4-  ;:'  +  ^z)  xyz.        40.  (««  +  ^'^  +  c)  a/^V. 

36.  (I  +  a  +  «*  +  a*)  «*•  41.  Qri/z  —  «/2;2^  —  zwx)  xyzw. 

37.  (;r*  —  .r*  +  .i'"  —  ^  +  1)  a^.      42.  (^*  +  ^V  +  ^'^^  +  '^«^)  -^y-^- 

38.  (a»  +  a^^  +  ab^  +  6«)  a*.  43.  (2  ahj  -Sbh-4:  c^w)  5  abc. 

44.  (1  -  a*6^  +  a^b^c  —  a^b^ed)  abed. 

45.  (4  a^x  -  3  b^y  +  2  ^r^")  5  cxy. 

46.  (8  a"//*  +  13  6V  +  6  c2^)(-  4  a6c^. 

47.  (a"  +  a"+^  +  a""^^  +  «"+*)  a. 

48.  C^**"  +  ^'^  +  P"  +  ^z")  ^>. 

49.  (c"-'  +  C-^  +  c'-i  +  c«)  c. 

50.  (d^-^  +  d^  +  d^+^  +  d^+^  d^. 

51.  (w^'-'  4-  w^-i  +  w"-"'  +  w""+')  w'. 

52.  (tw*'--^  -  w»'-^  +  m^'^^  —  ttT^)  mK 

53.  (a''-^  +  a"-^^"  +  a^^^^^  4-  ^""^')  ab. 

54.  (ar"-*  -  ;r"-»/-i  +  ^-y-«  -  3/"-^  ^V- 

55.  (a*'+=^  4-  «'"+'^'"  -  a=^"^'*+^  -  ^=^^+')  a'^r-. 

56.  (w''+^  —  ^"^+2^2  ^  ^-+3;^«  _  n^)  2^«- V+^ 

Multiplication  of  One  Polynomial  by  Another 

10.    We  have,  by  the  Distributive  Law  for  multiplication, 

{a  +  h  —  c){qc  —  y)  =  ax  +  bx  —  ex  —  ay  —  by  4-  cy. 
Hence,  to  multiply  one  polynomial  by  another^  multiply  eaeh  term 


MULTIPLICATION  97 

of  the  multiplicand  by  each  term  of  the  multiplier^  and  write  the 
algebraic  sum  of  the  resulting  partial  products. 

Ex.  1.    Multiply  2a:2  4-3a;-7by3a:-5. 

Two  arrangements  of  the  work  are  shown  below : 

Form  I  Form  II 

Check.     Let  a:  =  2. 


2x2+    3a;_    7 Multiplicand . 

Zx  —    5 Multiplier  .  . 

2x2+    3a;_    7  _  , 

3  X  -    5  .  . 

.  .  7 
.  .  1 

6x3+    9a:2_21x                 Partial 

-  10  x2  -  15  X  +  35          Products 
6ar8_       x2_36x  +  35    ..Product    . 

-  10x2-  15X  +  35 
6x8+    9a;2_21x 
.6x8-       a:2_36x  +  35  .  . 

7 
.  .  7 

0 

In  Form  I  the  first  and  second  rows  of  partial  prockicts  are  obtained  by 
multiplying  the  terms  of  the  multiplicand  successively  by  3x  and  by  —  5 
respectively. 

In  Form  II  the  arrangement  corresponds  to  the  one  adopted  in  arithmetic 
multiplication. 

Ex.  2.    Multiply     3x3  +  x2-4x  +  5byx2-x-3. 

For  convenience  in  performing  the  work,  it  is  usually  best  to  arrange  the 
given  polynomials  according  to  increasing  or  decreasing  powers  of  some 
letter. 

Arranging  both  multiplicand  and  multiplier  according  to  descending 
powers  of  x,  we  have : 

Check.     Let  x  =  2. 

Multiplicand 3x8+     a:2  -4x  +5       25 

Multiplier x2  -     x   -3 -    1 

p.   f.  1       (  using  x2  3  a:6  ^     a;4  _  4  ^.8  _|.  5  y2  _  25 


k 


Products  ■)  ^^'^"-  -  ^  -  3  x^  -     x8  +  4  X-   -    5  X 

using  -  3 -9x8-3x2   +  12x-15 

Reduced  Product ...      3  x^  -  2  x*  -  14x8  +  6x2  +    7  a;  _  15    _  25 

0 
11.  From  the  Distributive  Law  for  the  maltiplication  of  polyno- 
mials, it  appears  that  the  product  of  an  algebraic  sum  of  m  terms 
and  an  algebraic  sum  of  n  terms  will  contain  mXn  partial  products 
if  all  be  written  directly  without  the  collection  of  like  terms  or  the 
suppression  of  terms  which  mutually  destroy  each  other.  This 
principle  relating  to  the  number  of  terms  in  the  reduced  product 
will  sometimes  serve  as  a  check  upon  accuracy. 

7 


98  FIRST  COURSE   IN  ALGEBRA 

E.  g.  When  2x  -^3y  +  z  +  w  is  multiplied  by  a  +  6  —  c  there  will  be 
twelve  sepjirate  partial  products  in  the  result,  for  the  first  polynomial  con- 
tains four  and  the  second  contains  three  separate  terms. 

12.*  Detached  Coefficients.  When  two  expressions  are  both 
arranged  according  to  either  descending  or  ascending  powers  of 
some  letter,  the  work  of  multiplication  may  be  shortened  by  writing 
only  the  numerical  coefficients  of  the  different  terms,  writing  0  in 
place  of  each  missing  term. 

Ex.  3.     Multiply  3 x^  +  a:  -  4  by  2 a;^  -  3a;  +  5. 


Using  Literal  Factors. 

Using  Detached  Coefficients. 

Check.     X 

=  1 

3a:2+    X  -    4 

3      +1 

-    4 

.  0 

2x2  _  3a.  ^     5 

2      -3 

6      +2 

+    5 

A 

6x*  +  2a:«-    8x^ 

-    8 

0 

-9x«-    3a:2+12a; 

-9 

-    3      +12 

+  15x2+    5a. 

-20 

+  15      +5 

-20 

6x*-7x«+    4x2+17x- 

-20. 

6      -7 

+    4      +17 

-20 

(^)- 

.  0 

6x*-7a:»+    4x2+ 17a; 

-20. 

(2). 

.  0 

The  highest  power  of  x  in  the  product  is  x*,  and  since  the  terms  in  the 
multiplicand  and  multiplier  are  arranged  according  to  decreasing  powers 
of  X,  and  no  powers  are  missing,  the  remaining  powers  in  the  product  will 
follow  in  successive  terms. 

Hence  from  (1),  we  may  write  as  the  product  required 

6x^-7x8  +  4x2+ 17x- 20.  (2) 

By  the  use  of  detached  coefficients  the  labor  of  writing  all  the  literal 
factors  and  their  exponents  has  been  saved. 

Ex.  4.    Multiply  5  x*  -  6  x  +  3  by  2  x2  -  4. 

Supplying  the  "  missing  terms,"  it  may  be  seen  that  the  multiplicand 
and  multiplier  are  equivalent  to  5  x^  +  0  x2  —  6  x  +  3  and  2  x2  +  0  x  —  4 
respectively. 

The  process,  using  detached  coefficients,  is  shown  below : 

5  +  0-    6  +  3 

2  +  0-    4 
10  +  0-12  +  6 

-20-0  +  24-12 
10  +  0-32  +  6  +  24-12 

*  ThiB  Bection  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 

/ 


MULTIPLICATION  99 

The  term  containing  the  highest  power  of  x  in  the  product  is  obtained  by 
multiplying  hx^  by  2x2.     Accordingly  beginning  with  x^  we  may  write 
the  successive  terms  of  the  reduced  product  as  follows : 
10  xS  +   0  a;4  _  32  a:8  +  6  a:2  +  24  a;  -  12, 
that  is,  10a:«  -  32  x^  +  6a;2  +  24  a;  -  12. 

The  detached  coeflicients  may  be  used  for  a  numerical  check  as  in  Ex.  3. 

13.  A  polynomial  is  said  to  be  homog-eneous  if  the  sums  of  the 
exponents  of  all  of  the  letters  appearing  in  the  different  terms  are 
the  same. 

E.  g.  The  polynomial  a*  —  a%  +  a^^^  _  qX;&  ^.  ^4  \^  homogeneous  and  of 
the  fourth  degree  with  reference  to  a  and  6,  since  the  sum  of  the  exponents 
in  each  of  the  terms  is  4. 

The  polynomial  5  x^  +  2  x^\p-  —  a:*t/*  +  3  ari/^  —  -if  is  homogeneous  and  of 
the  sixth  degree  with  reference  to  x  and  y. 

Homogeneity  as  a  "Check  upon  Accuracy 

14.*  The  product  of  two  homogeneous  expressions  is  a  homogene- 
ous expression. 

For  if  the  homogeneous  multiplicand  is  of  the  m^^  degree,  every 
term  will,  by  definition,  be  of  the  w"*  degree.  Also,  if  each  term  of 
the  homogeneous  multiplier  be  of  the  n^^  degree,  then  since  each 
term  of  the  product  arises  from  multiplying  a  term  of  the  multipli- 
cand by  a  term  of  the  multiplier,  each  of  the  partial  products  must 
be  of  the  {m  +  w)***  degree ;  that  is,  the  product  must  be  a  homo- 
geneous expression  of  the  {m  -j-  iif^  degree. 

Although  the  proof  is  given  for  two  factors  only,  it  may  be  ex- 
tended to  include  any  number  of  factors. 

This  property  of  homogeneous  expressions  may  be  used  as  a  check 
upon  accuracy,  since  if  the  product  obtained  by  the  multiplication 
of  one  homogeneous  expression  by  another  is  not  homogeneous, 
then  we  know  at  once  that  there  must  be  some  mistake  in  the 
work. 

E.  g.  The  product  obtained  by  multiplying  a*  +  a%  +  ah^  +  ¥  by 
a^  ~  ab  +  b^  is  a  homogeneous  expression  of  the  fifth  degree,  and  may 
contain  terms  such  as  a^,  a%,  aW,  a%^,  ab\  and  b^.  It  cannot  contain 
terms  such  as  a*,  a^,  a^^,  ab^,  etc. 

•  This  section  may  be  omitted  when  the  chapter  is  read  for  the  flrpt  time. 


100  FIRST  COURSE   IN  ALGEBRA 


Perform  the  following 
results  numerically  : 

indicated 

multiplications,   checking  all 

1.  a  +  6 

^'  f+g 
f-g 

7.   27W  +  3w 
4?w  +  5?» 

2.   b  +  c 
b  +  d 

5.  F  +  >t 

k  +/ 

8.   6^-  \<dn 

Z.   c  —  d 
d-e 

6.   r^-r« 

r^  +  r* 

9.  a^  +  2ab  +  b" 

10.  («^-a6  +  ^>^(«  +  b). 

11.  (2;r*+ 5.r?/+  7?/^(2ar  — 3?^). 

12.  (2c*+ 3cc?  +  4i^(6c-5cr). 

13.  (w*  +  w^+  l)(tv*  -  w^  +  1). 

14.  («2  +  2  a  -  3)(a=*  +  «  —  6). 

15.  (5r»4- 4r-2)(3r2  — 2r-5). 

16.  (a^  -Sab-  b''){a'-  +  Sab  +  b''). 

17.  (^'  +  gV  +  y'W  -  gV  +  /). 

18.  (F  -  %  +  ?^^(2  F  -  3  A-z^  +  y=0- 

19.  Qi^  +  hhj  +  kf-^f){h-y). 

20.  (c?»-4c?^  + 36?+  l)(c?2-2<?+ 5). 

21.  (/-5/+  l)(2/+5f/+l). 

22.  (s«  — 3s*+ 2s2— l)(s^  — .9+  1). 

23.  (6>P  +  4F  +  2^+  l)(/5:^-A-2-l). 

24.  la^-ah-vb'-^  a-\-b-\-\){a-\-b-  1). 

25.  (1  —  27W  +  2w^2  — W2»)(l  +  2w  +  2^2  + wz*). 

Multiply 

26.  a^  -\-  y"^  +  z^  —  xy  —  zx  —  zyhy  X  +  y  +  z» 

27.  ^  +  ^V  +  a^f  +  ^y  +  ^^*  +  /  by  ar  —  y. 

28.  x""  —  xY  +  A^  -  xy  +  /2  |3y  ^2  ^  yj^ 

29.  «2^  -  6V  -  c«^  +  .^«  by  a^^c^  -  ab^c. 

30.  5a«^>'*-46V+ 5c»a»by4d;6  — 5^>c  +  4ca. 

31.  a%c  -  ab^c  +  a^>c2  by  ah^c^  -  a'^bc^  +  a%''c. 

32.  a6V  -  a6«c2  +  a'^bc^  by  a^Z^s^  _  a^c"  +  ««^*'c. 


MULTIPLICATION  101 

No  change  in  the  process  is  necessary  when  some  or  all  of  the 
coefficients  are  fractional. 

Multiply 

33.  ia'+ia+JbyJa  +  J.     36.  ^m^+ mn  +  ^7i^hy  ^m  + ^n. 

34.  i.r^-J^  +  ibyi^'-i.     37.  ^a^-^a'+la-lhyQa-j^. 

35.  W  —  iab+lb^hyia  +  ^b.  38.  ^-fa  +  fa^— f  «^by  2-5«. 

Ex.  39.   Find  the  product  of  4  a»+^  —  3  a"  +  V  rt«-i  and  5  rt«+i  —  2  ««. 

The  process  of  multiplication  is  not  affected  in  any  way  because  the 
letter  n  appears  as  an  exponent.  We  shall  consider  for  the  present  that  n 
represents  a  positive  whole  number.  The  separate  partial  products  are 
obtained  by  adding  the  exponents  of  the  like  factors  entering  into  them. 

E.  g.  The  first  terra  of  the  multiplicand  may  be  multiplied  by  the  fii'st 
term  of  the  multiplier  as  follows  : 

4a«+i  X  5an+i  =  20  a<«+i>+'n+i>  =  20a2«+2. 
The  remaining  partial  products  may  be  found  in  a  similar  way. 
Multiply 

41.  4a"+2  4-  3^»+i  +  2a''  +  1  by  a  -  1. 

42.  5er*''  + 4.r»"+ 3^2"  + 2^"by  2;r- 1. 


43. 

a-%^  -  a'^-'fj'  +  a'^-'b  by  a«6"  -  a%\ 

44. 

3  «'"+^  -  7  a"'  +  4  a"'-'  by  2  a"*  -  5  a'-^ 

45. 

^2n  +  ^«  4.  1  by  ^'2«  _  ^  _  1, 

46. 

af+y-^  +  af^y+'-  by  x^-hf'+^  -  ^^+y-\ 

47.  a:^"-^'  -  a-""  +  ^^"-i  by  ^""^  +  ^^""^  +  ^"'-^. 

48.  of-"  +  .z^*  +  of-'  by  .7f+''  +  eZ;"+*  +  af'+', 

49.  «"•+"  +  a"*  +  1  by  «"-"  —  a'"  +  1. 

15.    Removal  of  Parentheses.     Parentheses  may  be  removed 
by  applying  the  Distributive  Law  for  Multiplication. 

Ex.  1.     Simplify  6a  -  5{a  -  4  [3  +  2  (a-  1)]}. 

We  have,  6a  -  5{a  -  4  [3  +  2  (a  -  l)]}  =  6«-5|a-  4  [3  + 2a -2]} 

=  6a-5{a-4[l  +  2a]} 
=  6a-5|a  — 4-8a} 
=  6a-5{    -4-7a} 
=  6  a  +  20  +  35  a 
=  41  a +  20. 


102  FIRST  COURSE  IN   ALGEBRA 

Exercise  VII.   5 
Simplify  each  of  the  following  expressions  : 

1.  1  +  2(1  +3[1  4-4(1  +  5^')]}. 

2.  a:~{(j;-z)-la:  +  2/-z-2(^-ij  +  z)l}. 

3.  2  +  2{2  -  2  [2  +  2  (2  -  2ar)]} 

4.  (.r  +  1)  -  2{(^  +  2)  +  3  [(^  +  3)  -  4  (;r  +  4)]}. 

6.  5{4  [3  (2  +  a)]}-  5{-  4  [-  3  (2  -  a)]}. 

7.  7{^  _  4  [6  _  4  (^  +  ^)]}_  6{6  -  4[6  -  2  (6  -  i/)]}. 

8.  a{l  +  6[1  +c(l  +d)^}-d{i  +  c[l  4-^(1  +«)]}• 

9.  a{b  —  c[a  —  b  (a  +  b  +  c)  —  {b  +  a)]  -  c  —  0}. 
10.  ^^^  -  ^{/>  +  cla  (b  -  c)  +  b  (c  -  a)  +  c  (a  -  b)]}. 

Standard  Identities 

16.  Special  Products.  Just  as  in  Arithmetic  we  find  it  neces- 
sary to  commit  to  memory  the  Multiplication  Table,  so  in  Algebra 
certain  products  occur  so  frequently  that  it  is  important  to  mem- 
orize them. 

17.  A  polynomial  expression  is  said  to  be  an  expansion  of  a 
second  polynomial  expression  if  it  is  obtained  by  raising  the  second 
expression  to  some  power. 

E.  g.   The  expansion  of  (a  +  6)2  is  a^  +  2  a6  +  b^. 

18.  Square  of  a  Binomial  Sum. 

Theorem  I.    {a  +  b)^  =  a:^  +  2ab  +  b^. 

The  sqtuire  of  a  binomial  sum  is  eqiml  to  the  square  of  the  first 
terrriy  increased  by  twice  the  product  of  the  two,  plus  the  square  oj 
the  second. 

Check.  Let  x  =  Z. 
Ex.1.    (x  +  4)2      =x2  +  8a;  +  16.  49  =  49. 

Check.  Let  m  =  n  =  4. 
Ex.  2.    (m  +  w)2    =  m2  +  2  mn  +  n^.  64  =  64. 

Check.  Let  x  =  3,  i/  =  2. 
Ex.3.    (2a;  +  3i/)2=(2a:)2  +  2-2a:-37/  +  (3i/)2  144=144. 

=  4x2+  i2x?/  +  9i/2. 


STANDARD   IDENTITIES 


103 


Mental  Exercise  VII.    6 


Expand  each  of  the 

1.  (a  +  3)^ 

2.  (b  +  4)1 

3.  (c  +  6)^ 

4.  (c  +  1)\ 

5.  (5  4-  d)\ 

6.  (8  +  h)\ 

7.  {k  +  9)^ 

8.  {m  +  10)'^. 

9.  (11  +  n)\ 
10.  [x  +  12)^ 

(13  +  yy. 
(z  +  14)^. 

(15  +  wy. 

{2a+iy. 

15.  (3  6+  l)'^. 

16.  (4c+  1)=^. 


11. 
12. 
13. 
14. 


following  binomial  sums  : 

17.  (5d+  ly.  33. 

18.  (4^+  3)^^.  34. 

19.  (5^  +  4)^  35. 

20.  (3^  +  5)''.  36. 

21.  (6^+  7)^  37. 

22.  (5  +  8  hy.  38. 

23.  (9  +  6^)'*.  39. 

24.  (7  +  9  my.  40. 

25.  (10^+  3)=^.  41. 

26.  (4a  +  5  by.  42. 

27.  (3c+  Idy.  43. 

28.  (6/i  +  10  ky  44. 

29.  (5w+  13  ny.  45. 

30.  (2/'+  llijy.  46. 

31.  (16  5+  3wy.  47. 

32.  (12;r+  5ijy.  48. 


(7^  +  20  zy. 

(18  A:  +  10  ny. 
(19a  +  2c)^ 
(3  6+  18  i)^. 
(20  c  +  5gy. 
(10  6+  16  ^)^ 
(14  6^+5^)=*. 
(12 /i+  11  r)''. 
(21.9+  2v)^ 
(8  ab  +  5)^ 
(9  .r?/  +  10)^ 
(11 «  +  4  6c)'*. 
(l,T  +  4yzy. 
(6  a6  +  5  cd)^. 
(10  XIV  +  9ijzy. 
(15ac+  5  6c0'. 


19.     Square  of  a  Binomial  Difference. 

Theorem  II.    {a-b)^  =  a^  -2ab  +  b\ 

TTie  square  of  a  binomial  difference  is  equal  to  the  square  of  the 
first  term,  diminished  by  twice  the  product  of  the  two,  plus  the  square 
of  the  second. 


\ 


Check.     Let  x  =  Q. 

4  =  4. 

Check      Let  z  =  2. 

1  =  1. 

Check. 

Ex.3.    (3x-5y)==(3x)2   -2  •3a;-5i/  +  (5i/)2        Let  a:  =  4,  y  =  2. 

=  9  a;2  -  30x1/  +  25  y^.  4  =  4. 


Ex.1,    (a: -4)2       =  a;^  -  8a:  +  16. 

Ex.  2.      (3-2)2  =9_62+2;2. 


Mental  Exercise  VII.  7 
Expand  each  of  the  following  binomial  differences  : 
1.  {a-^y.  3.   {c-^y.  5.   {b-ey. 


2.  (6  -  ly. 


4.  (4:-d)' 


6.  (6  -fy. 


104  FIRST  COURSE  IN  ALGEBRA 

7.0-  9)^  21.   (8  ^  -  1)^  35.  (8  ;r  -  7  z)\ 

8.  (10  -  h)\  22.  (3  k  -  5)^.  36.  (14 1  -  4.yf. 

9.  \h-\\y.  23.  (4^-7)^  37.  {lOq-Szf. 

10.  (12  -  my.  24.  (6v-5)2.  38.  (6r-  lly)«. 

11.  (w  -  14)''.  25.  (7  «^  —  9)^.  39.  (15  m  —  ^sf. 

12.  (16-r)\  26.  (8-65)'.  40.  (I2  7i-20y)l 

13.  («  -  17)^  27.   (11-4  ty.  41.  (17  g-^  rf. 

14.  (18  -  0*-  28.   (13  r  -  3)^.  42.  (2  ab  -  18)'. 

15.  [x  -  19)^  29.  (2  5  -  15)^  43.  (5-4  cg)\ 

16.  (2  a  -  \y.  30.  (4  5'  -  8)^  44.  (5  ab  -  10  cd)\ 

17.  (76- I)'*.  31.  (9a -36)2.  45    (9 ^n^;  -  4 yz)'*. 

18.  (9  c-  \y,  32.  (6  c  -  7  oT)"-  46.  (8  mx  -  ^  nyf, 

19.  (1-4  d)\  33.  (5  A  -  12  r)'.  47.  (2  ay  -  ^  bx)\ 

20.  (1-5  ey.  34.  (9 jt>  -  7  qf.  48.  (7  erf  -  6  M)'. 

20.  Multiplication  of  the  Sum  of  Two  Terms  by  their 
difference. 

Theorem  III.    (a  +  h){a  -b)  =  a^-  bK 

The  product  obtained  by  multiplying  the  sum  of  two  terms  by  their 
difference  is  equal  to  the  difference  of  the  squares  of  these  terms. 

Ex.1.    (x  +  5)(x-*5)  =a;2-25. 

Ex.2.    (9  +  m)(9-m)  =  81  -  m^. 

Ex.3.    (5x  +  6y)(5x-6i/)  =  25a:2- 361/2. 

Let  the  student  check  each  of  the  examples  above. 

Mental  Exercise  VII.    8 
Obtain  the  following  products  : 


1.  (x  +  y)(x- 

-y)- 

10.  {k  +  9)(^  -  9). 

2.  (c  +  k)(c  - 

■  k). 

11.    {n+g)(\l-gy 

3.  (r  +  M7)(r- 

-w). 

12.  {h  +  13)(^  -  13). 

4.  (m  H-  qXm 

-q)' 

13.  (14  +  ^)(14-4 

5.  (a  +  z)(a  - 

-z). 

14.  {m  +  \l)(m  —  17). 

6.  (x  +  3)(x  - 

-3). 

15.  (16  +  n){U  —  n). 

7.  (l/  +  4:)(y- 

-4). 

16.  (2a+  l)(2a- 1). 

8.  (5  +  z)(5  - 

-z). 

17.  (3  6+l)(3  6-l). 

9.  {e  +  wxe- 

-w). 

18.  {c-\-2d){c'-2d). 

STANDARD   IDENTITIES 


105 


19. 

:c  +  id)(c~id). 

35.  0 

20. 

[a  +  bb)(a-'ob). 

36.  0 

21. 

\a  +  8c)(a- 8  c). 

37.  0 

22. 

(4a  +  36)(4a-36). 

38.  0 

23. 

05c+7^)(5c-7^). 

39.  ( 

24. 

(8^'  +  3^«^)(8^-3^^')• 

40.  0 

25.  ( 

;9;^^-  ll;^)(9?i-  lis;). 

41.  0 

26.  ( 

'12  m  +  8r)(12m-  8r). 

42.   ( 

27.  ( 

;7^+  15  0(7^-15  0- 

43.   ( 

28.   ( 

;i3w  +  90(13w-9s). 

44.  0 

29.  < 

;i5^  +  12;2)(15(?-12  2:). 

45.   ( 

30.  { 

;i7^+  6v)(17c?-6v). 

46.   C 

31.  ( 

[\Hn  +  3s)(18w-  3.S'). 

47.  (] 

32.  { 

;iOjt>+  16  0(10;^- 16  0- 

48.  (] 

33.   ( 

;9^4-  20r)(9  5'- 20r). 

49.   (] 

34.  ( 

;i9a  +  5A)(19a- 5^). 

50.  (] 

21.    Square  of  a  Polynoiiiial, 

nomial  consiatin;?  of  three  terms  : 


(6c+  \2g){ioc—  12  g). 
(45+  14c)(4  5-  14.  z), 
(Sab  +  5)lsab  —  6). 
(2cfi?+  9)(2cc?  — 9). 
(7  +  &my){l  —  ^my), 
(8  +  9;2^)(8  -  9??c). 
(4  6?^+  5  c)(4  ab  —  5  c). 
(10  .r+  7y;:)(10^—  1  yz). 
(11M-+  div)(nbk  —  ^w). 
(Sga;  +  l{)ky)(Sga:  —  10  hy). 
{I2am  +  rdbn)\uam  —  Vdbn). 
llabc  +  Sd)(labc-Sd). 
(14.2'+  l\yzw)(14:a'—llyzw). 
(15^^^+  10cc?)(15a^-  10c<^). 
(16^>c  +  16m7i)(Ubc  —  16mn). 
{llmn+  18pq)(nmn  —  lSpq), 

Consider  the  square  of  a  poly- 


(a  +  b  +  cy=  (a-\-  b  +  c)(a  +  b  ■\-  c) 
=  (a  +  b  +  c)a 
+  (a  +  b  +  c)b 
+  (a  +  b  +  c)c. 

From  the  arrangement  of  the  work  above  it  appears  that  each 
partial  product  obtained  by  multiplying  any  one  of  the  terms  in 
parentheses  by  the  factor  outside  must  be  of  the  second  degree. 
The  only  possible  terms  which  can  arise  in  this  way  will  be  those 
which  are  the  squares  of  the  given  letters,  such  as  a^,  b\  and  c^  and 
those  which  are  the  products  of  all  possible  pairs  of  the  letters,  such 
as  ab,  ac,  and  be. 

By  examining  the  identity  above  it  may  be  seen  that  a^  will  occur 
but  once,  that  is,  as  a  result  of  the  multiplication  in  the  first  line ; 
b"^  will  occur  but  once,  that  is,  as  a  result  of  the  multiplication  in 
the  second  line ;  c^  once  only,  and  that  from  the  multiplication  in 
the  third  line. 

Furthermore,  it  appears  that  any  product  such  as  ab,  of  two  dif- 
ferent letters,  will  occur  twice  and  twice  only. 


106  FIRST  COURSE   IN  ALGEBRA 

Thus,  ah  will  arise  from  the  multiplication  in  the  first  line  and 
also  from  that  in  the  second  line ;  ac  from  that  in  the  first  and  third 
lines  ;  be  from  that  in  the  second  and  third  lines. 

The  reasoning  employed  above  may  be  extended  to  include  a 
chain  of  additions  and  subtractions  containing  any  number  of  terms. 

Hence  we  have 

Theorem  IV.    (a  +  &  +  c)2  =  a^  +  6^  +  ^^  +  2  a&  +  2  ac  +  2  6c. 

The  square  of  any  polynomial  is  equal  to  the  sum  of  the  squares  of 
the  different  terms^  Increased  by  twice  the  product  qf  each  term  and 
every  term  which  follows  it. 

22.  The  square  of  a  polynomial  may  also  be  obtained  by  suitably 
grouping  the  terms  and  then  applying  directly  the  theorem  for  the 
square  of  a  binomial. 

E.  ^.   Find  the  square  of  a  +  6  +  c. 

Reganling  a  +  6  in  the  polynomial  as  a  single  term,  we  may  write 
a  +  6  -f  c  =  (a  +  6)  +  c. 

Hence,  [(a  +  6)  +  cf  =  (a  +  6)2  +  2  (a  +  6)  c  +  c^ 

=  a2  -I-  2  a6  +  62  4-  2  ac  +  2  k  +  c2 
=  a2  4-  62  +  c2  +  2  a6  +  2  ac  +  2  6c. 

Ex.  1.    Find  the  square  of  (a  +  3  6  +  5  c  +  7  rf). 

(a  +  36H-5c  +  7rf)2  =  a2  +  (36)2     +(5c)2+  (J  d)'^ 

Multiplying  2  a  by  following  terms    .    +  2  a  (3  6)  +  2  a  (5  c)  +       ^a{l  d) 

Multiplying  6  6  by  following  terms +  2  (3  6)(5  c)  +  2  (3  6)(7  rf) 

Multiplying  10cby7ci     +2(5c)(7rf). 

=  a2  +  9  62  +  25  c2  +49  d^ 

+  6a6  +  10  ac+  lAad 

Check,  a  =  6  =  c  =  (Z  =  2.  +  30  6c  +  42  6f? 

1024  =  1024.  +  70  cd. 

Ex.  2.    (5x  -  22/ -  42)2  =25a:2  +  4i/2  + 16^2  _  20x1/ -40a;;3+  16^2!. 

Check,     a:  =  4,1/  =  3,2  =  1.     100=100. 

Mental  Exercise  VII.     9 
Expand  each  of  the  following  polynomials  : 

1.  (a  +  26  +  3c)'.  4.  (4^  +  7?/  +  2z)\ 

2.  (4c  +  ^  +  bej  5.  (5«  -  b  +  2c)'. 

3.  (6^  +  3  ?*  +  w)\  6.  (4^  +  33/  -  zj\ 


STANDARD   IDENTITIES  107 

7.  (6a  +  2b-  r>  cy.  14.  (a-  2b  +  c  -  5  df. 

8.  (9(1  +  2b+  ly.  15.  (1  —a  —  b  —  c)'l 

9.  (7^-.^y+  2)^.  16.  (2a  +  Sb  +  4c  +  Sofjl 

10.  ((f  +  b  +  c+  ly.  17.  (a-b~c-d-  ef. 

11.  (^  +  y  -  ;^  -  1)1  18.  (a  -  2  6  +  c  -  3  ^  +  f?)2. 

12.  (a-b^-c-2)\  19.  (4.r  +  2?/- ;^  +  3«^^- 5)^ 

13.  (;6x  —  y  +  2z  —  wf,  20.  (^- 23/ +  3<^  — 2t«;  +  1)'». 

23»  Product  of  Two  Binomials  in  which  the  First  Terms 
are  Equal  and  the  Second  Terms  are  Unequal. 

Theorem  V.    {oc  -\-  a){3c  +  b)  =  x!^  +  {a  +  h)  oc  +  ab. 

The  product  of  two  binomials  having  eqtial  first  terms,  but  unequal 
second  terms,  is  equul  to  the  square  of  the  common  first  term,  increased 
by  the  product  of  the  algebraic  sum  of  the  second  terms  and  the  first 
term,  plus  the  product  of  the  second  terms. 

Ex.  1.  Multiply  (x  +  2)  by  (x  +  3). 

The  two  binomials  havH  equal  first  terms,  x,  but  unequal  second  terms, 
2  and  3.     Hence,  we  may  write  : 

(z  +  2)(x  +  3)  =  x2  +  (2  +  3)x  +  2  -3  Check.  a:  =  2. 

=  a;2  4-5a;  +  6.  20  =  20. 

Ex.2.    (2x  +  5)(2x  +  6)  =  (2x)2+(5  +  6)2x  +  5  •  6  Check.  a;=:2. 

=  4a;*  +  22iC  +  30.  90  =  90. 

Ex.  3.        (a;  +  7)(x  -  2)  =  x^  +  (7  _  2)  X  +  7  (-2)  Check,  a;  =  3. 

=  x2  +  5x-14.  10=10. 

Ex.4.  (3a  +  6)(3a-26)  =  (3a)2+(6-26)3a  +  H-20  Check,  a  =  3, 

&  =  4. 

=  9  a2  _  3  a6  -  2  6^.  13  =  13. 


Mental  Exercise  VII.   10 

Obtain  the  following  products 

: 

1.    (a  +  2)(«  +  1). 

7.    (^+6)(^  +  4). 

2.    (.r  +  3)(;r  +  2). 

8.    (^+6)(;r  +  8). 

3.    (x  +  ^)(x  +  1). 

9.    (x  4-  i)(x  +  2). 

4.    (^_4)(^-5). 

10.    (^+  7)(^-5). 

5.    («  +  5)(a  +  2). 

11.    (^+7)(^+6). 

6.    (x  +  h)(x  +  3). 

12.    (.r  +  8)(^-3). 

108  FIRST  COURSE  IN  ALGEBRA 

13.  (ic  +  8)(;i- +  7).  37.  0/ +  U)(7/ -  8). 

14.  (d-  +  9)(ii-  -  2).  38.  (m  +  17)(w  ~  2). 

15.  (m  +  d)(m  —  6).  39.  («  -  10)(a  -  19). 

16.  (a;- 10)(.«-2).  40.  (/^  +  18)(A  -  10). 

17.  (x  -  3)(x  +  10).  41.  (a  —  20)(«  —  15). 

18.  (a;-5)(a;+  11).  42.  (a;  -  20)(a;  -  5). 

19.  (x  -  2)(:x  -  12).  43.  {m  -  8)(m  +  20). 

20.  (x  -  Ui)(x  +  8).  44.  (a;  -  16)(a;  +  20) 

21.  Ix  +  U)lx  -  4).  45.  (a  +  15)(«  +  14). 

22.  (a;  +  4)(a;  +  13).  46.  (b  -  22)(6  -  3). 

23.  (a  +  d)(a  +  10).  47.  {a  +  25)(rt  +  8). 

24.  (a  +  11)(«  +  7).  48.  (c  +  20)(c  +  30). 

25.  (w-ll)(w+10)  49.  (2a+3)(2«  +  5). 

26.  (a;  -  12) (a;  +  1).  50.  (4  6  -f  5)(4  b  +  1). 

27.  («r  -  13)(a  -  2).  51.  (6c  +  l)(6c  -  3). 

28.  (a  +  3)(a  +  14).  52.  [l  d  -  2){1  d  -  4). 

29.  {a  +  I2){a  -  1).  53.  (5  ^  +  8)(5  //  -  6).  . 

30.  (c  +  13)(c  -  3).  54.  (8  ^  -  11)(8  h  -  1). 

31.  (a-3)(a-16).  55.  (3^^  -  1)(3A;  -  19). 

32.  (6  +  15)(6  +  4).  56.  (9  w  -  2)(9  w  -  4). 

33.  (a  +  10)(a  +  16).  57.  (3«  +  5)(3w  -  8). 

34.  («  +  5)(a+18).  58.  (11  a;  -  9)(11  a;  +  2). 

35.  (a  +  19)(a-2).  59.  l\2y  +  l){l2y  -  b). 

36.  (c--3)(c- 17).  60.  (10;s-3)(10;s  +  4). 

24.  The  Product  of  Two  Binomials  of  the  forms  aic  +  h 
and  ex  +  d,  the  first  terms  of  wliicli  contain  a  common  fac- 
tor X,  may  be  found  by  applying  the  Law  of  Distribution  for 
Multiplication. 

Theorem  VI.    {ax  +  h)[cx  +  d)=  acx^  +  {ad  +  he)  x  +  hd. 

That  is,  the  product  of  two  binomials  of  the  types  ax  +  b  and 
ex  +  d  is  equal  to  the  product  of  the  first  terms  of  the  binomials, 
increased  by  ths  sum  of  the  ^^  cross  products"  plus  the  product  of  the 
second  terms. 

Such  products  as  adx  and  bcx,  shown  above,  are  commonly  called 
cross  products,  since  when  the  given  expressions  are  arranged  as 
multiplicand  and  multiplier  in  the  vertical  form  for  multiplication, 


STANDARD   IDENTITIES 


109 


in  order  to  obtain  these  products  we  must   cross   over  from   one 
column  to  another  as  suggested  below  : 

ao!  +  b 

X 

cw  ■}-  d 

hex  +  cidiC  or  {ad  +  he)  x. 
The  remaining  terms  of  the  product,  acx'^  and  bd^  are  obtained  by- 
multiplying  together  terms  which  are  in  the  same  columns. 

Ex.  1.    Multiply  (2  a  +  5)  by  (3  a  +  7).  Check,     a  =  2 

(2a  +  5)(3a  +  7)=  Ga^  +  29a  +  35.  117  =  117. 

25.    Products  of  expressions  of  the  types  ax  +  bij  and  ex  +  dy 
may  also  be  found  by  reference  to  the  Theorem  above. 

Check.  X  =  y  =  2. 
Ex.  2.  (3  »  +  8  y)(o  x-2y)  =  15  x^  +  .34  xy  -  16  ?/.  132  =  132. 

Mental  Exercise  VII.     11 
Obtain  each  of  the  following  indicated  products: 

I4c?+  11)(4<^+ 3). 
20^  -f  17)(5/^-4). 
13  a—  lS)(2a-  8). 
7g+12)(4^-7). 
6  5r+7)(ll^-10). 
12c- 13)(5c+  7). 

ldk+  l)(3k-  1). 
17  c  — 4)(3c—  10). 

13  b+  9)0'>^>-4). 
18^  +  7)(8^  — 3). 
15^-ll)(0^+5). 
16  m  +  9)(r)7n  +  3). 

14  71+  13)(7w- 6). 
')r  +  3s)(4r  +  Is). 
dp  —  4^)(5j9  —  2q). 

12  6  +  7d)(Sb  -  r)d). 

15  c-  13^)(4c-f  3^). 
14  A  +  17^)(3A-4^). 

16  m  —  dw)(Hm  +  6w). 

13  r+  dz)(lr  —  4z). 


1. 

(3«  +  2)(«+  1). 

22.  ( 

2. 

(2b+  l)(3  6  +  4). 

23.  C 

3. 

(4c+  l)(2c  +  3). 

24.  ( 

4. 

(Dd+l)(d+l), 

25.  ( 

5. 

(<7  +  2)(6i7+5). 

26.  (( 

6. 

(lh-\-  S)(2h+  1). 

27.  ( 

7. 

(8^+5)(^-l). 

28.  0 

8. 

(9  m  — 2)(m  +  1). 

29.  ( 

9. 

(6/i-  l)(4w  +  3). 

30.  ( 

10. 

(5r-4)(2r-.3). 

31.  ( 

11. 

(7,_9)(,_4). 

32.  ( 

12. 

(ll;r-8)(3^+2). 

33.  ( 

13. 

(l0  7y-7)(57/-4). 

34.  ( 

14. 

(l2^+5)(7^-3). 

35.  ( 

15. 

(13  w;  4-  G)(r)iv  —  2). 

36.  0 

16. 

(Ss-  15)(3  5+  5). 

37.  (1 

17. 

(92^+  ll)(4^-5). 

38.  ( 

18. 

(I5i;+  14)(5t'  — 4). 

39.  ( 

19. 

(6^+  ll)(2k-f)). 

40.  ( 

20. 

(\()a  +  \[))(2a  +  1). 

41.   ( 

21. 

(16  6- 9)(5  6  +  3). 

42.  ( 

110  FIRST  COURSE  IN  ALGEBRA 

26.  The  product  of  two  polynomials  consisting  of  the  same  terms 
arranged  in  the  same  order,  but  in  which  the  signs  of  one  or  more 
pairs  of  corresponding  terms  differ,  may  often  be  found  by  so 
grouping  the  terms  as  to  make  the  polynomials  appear  as  the  sum 
and  difference  of  the  same  combinations  of  terms. 

Ex.  1.    Multiply  x  +  y-{-zhyx  +  y  —  z. 

The  terms  of  the  polynomials  may  be  grouped  as  follows : 

x  +  y-{-z  =  {x  +  y)-{-z  and  x -{•  y  —  z=  (x  +  y)  —  z. 
Hence,  we  may  write  [(x  +  y)  +  2]  [(x  +  y)  —  z]  =  x'^  -^  2  xy  +  y^  —  zK 

Ex.  2.    (x  +  y-\-  3)(x  _  y  +  3)  =  (a:  +  3)2  _  y2 

=  x2  +  6x  +  9-y2. 

The  student  should  check  the  examples  above  numerically. 

Mental  Exercise  VII.     12 
Obtain  each  of  the  following  products  : 

1.  (771  +  n  +  q){m  -^  n  —  q).         7.  (.r  +  y  +  z){x  —  y  —  z). 

2.  (a  +  c  +  4)(a  +  c  —  4).  8.  {m -\-  n  +  \){m  —  n—  l). 

3.  {b  +  d-\-  2)Q)  +  6?  -  2).  9.  (^  +  c  +  3)(6  -  c  -  3). 

4.  {a  +  h+  \)ia  —  b -{■  \).         10.  {a  +  b  -  c)(a  —  h  +  c). 

5.  (2  +  c  +  d){2  —  c-\-d).  \l.  {x-\-y  —  z){x  —  y  —  z). 

6.  (a  +  6  +  c){a  —  b  —  c).  12.  \m  +  n  —  5)(w  —  w  +  5). 

Ex.  13.     (a  +  6  +  3)(a  +  6  +  2)  ^  (a  +  6)2  +  5  (a  +  6)  +  6 

B  rt^  +  2a6  +  6«  +  5 a  +  5 6  +  6. 

Ex.  14.    (x  -  2/  +  7)(a:  -y  -  A)  =  {x  -  yY  +  3{x  -y)  -  '2Q 

=  x2  -  2  a;?/  +  2/'  +  3  X  -  3  2/  -  28. 

15.  {m  +  n  +  2)(m  +  w  +  1).  21.  (a  —  b  +  6)(a  —  b  -  2). 

16.  lj^  +  y  +  4)(.2r  +  y+  5).  22.  (b  -  c  +  9)(6  -  c  -  5). 

17.  («  +  6  +  6)(a  +  6  4-  3).  23.  (c  —  ;r  —  3)(c  —  ;r  —  4). 

18.  (6  +  c+7)(64-c+2).  24.  (d  -  y  -  l)(d  -  y  -  2). 

19.  (^  —  y  +  4)(.r  —  y  +  1).  25.  (m  —  q—  10)(m  —  g  —  6). 

20.  (y  —  ;2  +  o)(^  —  ;:;  +  3).  26.  {m  —  w—  ll){m  —  w—l). 

Ex.  27.    (a  +  6  +  c  +  d){a  +  h  -  c  -  d)  =  {a  +  hy  -  {c  +  dy 

=  a2  4.  2  a6  +  62  _  c2  _  2  c(?  -  (^2. 

28.  {a'^-y-\-z-\-w)ix-\-y—z—^v).    31.  («  — ^;  +  c--^(«  — ^>  — c  +  cT). 

29.  la-\-b-\-c-{-l){a+b—c—\).       32.   (;r— ?/  +  2:— w)(^— y— ^+^')' 

30.  (;r+y+;5+3)(^+2/->2-3).      33.   (a— 6+c— 2)(a-6— c+2). 


STANDARD   IDENTITIES  111 

Powers  of  a  Binomial. 

27.  The  Binomial  Theorem.  The  student  should  obtain  the 
following  identities  by  actual  multiplication:    q]^qq^    L^^  ^  _  ^  _  1^ 

(a-\-by  =  a  +       b 2^  =    2.    ' 

la  +  by  =  a^+2ab  +       b^ 2^  =    4. 

(a  +  bf  =  (/  +  'Sa^b  +  3  «^2  +        b^ 2«  =    8. 

(a  4-  by  =  a*  +  4rt«/>  +  Qa%''  +  4«6«  +  ^^ 2^  =  16. 

It  will  be  well  for  the  student  to  extend  this  set  of  identities  as 
far  as  the  tenth  or  twelfth  powers  of  the  binomial  {a  +  b). 

By  inspection  of  the  identities  above,  we  shall  discover  certain  laws 
of  coefficients  and  exponents  which  will  be  found  to  hold  true  for 
any  positive  integral  powers  of  any  binomial.  (See  Chap.  XXVIII. 
§§2,3). 


112  FIRST  COURSE   IN  ALGEBRA 


CHAPTER  VIII 

DIVISION  OF  INTEGRAL  ALGEBRAIC   EXPRESSIONS 
(For  Definitions,  see  Chapter  V.) 

1.  When  au  expression  A  can  be  produced  by  multiplying 
together  two  others,  B  and  (7,  then  B  and  C  are  called  factors  of  A. 
The  expression  A  is  said  to  be  exactly  divisible  by  B,  and  also  by  C. 

In  multiplication  we  are  given  two  factors  to  find  their  product, 
while  in  division  we  assume  a  given  expression  to  be  the  product  of 
two  factors,  and,  having  one  of  them,  our  problem  is  to  find  the 
other. 

2.  Since  the  operation  of  division  is  the  inverse  of  that  of  multi- 
plication, it  follows  that  if  we  multiply  the  quotient  by  the  divisor 
we  shall  always  obtain  the  dividend. 

3.  To  divide  one  power  of  a  base  by  another  power  of  the  same 
base,  we  may  apply  the  following 

Law  of  Exponents:  The  qiwtient  obtained  by  dividing  any 
power  of  a  given  base  by  a  lower  power  of  the  same  base  is  a  power 
of  that  base.  Its  exponent  is  found  by  subtracting  the  exponent  of 
the  divisor  from  the  exponent  of  the  dividend. 

That  is  ^   ^'-^   «--«"  =  «--".  if^>^. 

If  m  and  n  are  positive  whole  numbers,  we  may  from  the  definition  of  a 
power  write  the  following : 

(i.)   m  >  w. 

,- TO  factors >        , n  factors . 

flwi  -1.  a"  =  (axaxaxax xrt)-f(axaxax  •  •  •  xa) 

, — (m—n)  factors — >    , n  factors ^    , n  factors- 


=  (axaxax  •  •  •  xa)x(axaxax  •  •  •  xa)-r(axaxax  •  •  ■  xa) 

, — (m—n)  factors — n 
=  {axaxax  •  •  •  xa) 
=  a"*~",  by  notation. 


DIVISION  113 

(ii.)   m  <  71. 

, m  factors v        r— »  factors -^ 

a-n  -i-  a"  =  (axftxrtx  •  •  •  Xfl')-=-(axaxaxax xa) 

, m  factors v   , m  factors >    - — {n—m.)  factors — v 

=  (axaxax  •  •  •  y.ci)-^{ay.ay.ay.  ■  •  •  xa)x(axaxax  •  •  •  xa) 

, (n— nt)  factors > 

=  1  -f  (axaxaxax xa) 

=  1  -^  a"""*,  by  notation. 

4.  According  as  the  exponent  of  one  power  is  greater  than  or  less 
than  the  exponent  of  another  power,  the  first  power  is  said  to  be 
bigher  or  lower  than  the  second. 

E.  g.  a^  is  ta  higher  power  than  a^  or  a^;  but  it  is  a  lower  power  than 
a^  or  a^. 

5.  It  will  be  shown  in  a  later  chapter  that  the  Law  of  Exponents 
may  be  applied  whenever  the  exponents  are  positive  or  negative, 
integral  or  fractional  numbers.     (See  Chapter  XIX.  §  4.) 

One  power  divided  by  anotber  power  of  tbe  same  base. 

6.  The  quotient  resulting  from  the  division  of  one  power  of  a 
base  by  another  power^of  the  same  base  may  be  found  by  applying 
the  Law  of  Exponents. 

Ex.1.  a« -r  a*  =  rt*-*  =  a2. 

Ex.  2.  W  -^  56  =  /,9-5  =  /,4. 

Ex.  3.  (-  x)8  -1-  (-  xY  =  (-  a:)»-«  =  (-  x^. 

Since  all  even  powers  of  negative  bases  are  positive  numbers,  we  have 

Let  the  student  check  the  above  results  numerically. 

Mental  Exercise  VIII.     1 

Express  the  following  quotients  of  different  powers  of  the  same 
bases  as  single  powers  of  positive  bases  : 


I. 

8«  ^  3^. 

8.  2"  -f-  2\ 

15.  (-2)i'>^(-2)^ 

2. 

2'  --  2*. 

9.  3^^  -f-  3^. 

16.  a«  -^  a\ 

3. 

3«  --  3^ 

10.  4^^  ^  4^ 

17.  1/'  --  /A 

4. 

G«-f-6. 

11.  6^  -^6^. 

18.    C"^  -r  g\ 

5. 

5'  -  5^ 

12.   (-3)«-(- 

-8)^ 

19.    h''-rh\ 

6. 

4«  --  4«. 

13.  (-6)«-(- 

-6)^ 

20.  o''-^(f\ 

7. 

V  -r-  1\ 

14.  (-2)«-f-(- 
8 

-2). 

21.  m^'-^m'^ 

114  FIRST  COURSE  IN  ALGEBRA 

22.  k'^  -^  X;".  43.  a^'+^  -4-  a.  64.  ?/»«+'•  -f-  7/'-=^. 

23.  (-«)"  -T-  (r-7iy.  44.  ^''^^^  -J-  6.  65.  z''''^'^  ^  z^^-". 

24.  (-ry^-f- (—?♦)*.  45.  c*'^+^-i-c^  66.  a'*+^'^  ^  a^^^'^ 

25.  (-a?)"-f-(— ;r)".  46.  d^'"+« -^  (f.  67.  ^4^+'^  ^  ^8x+5^ 

26.  (-i/)"-^(-^)".  47.  /*"+*  -^  /i".  68.  c«"'+^"  -7-  c*'"+2« 


27. 

i-zr-^i- 

-^)^ 

48. 

^m+n  ^  ^n 

69. 

fPd+8  _j_  ^d+8^ 

28. 

a"*  -T-a. 

49. 

a^"'+^--a'». 

70. 

^+6+c  ^  ^ 

29. 

b''  ^  b\ 

50. 

^4«+7  ^  ^2«^ 

71. 

ym^^Z^yn 

30. 

c'^-rc. 

51. 

c^'-2  -^  c«'-. 

72. 

^r+.+6  ^  ^6^ 

31. 

(P'-r-d". 

52. 

^-+2  ^  ^x^ 

73. 

«»•+«+'•  -f-  «'"+''. 

32. 

W"  -^  7W*. 

53. 

a^r+2  ^  ^r+1 

74. 

ix+y^-r  _:_  Jf+z 

33. 

a*"  -^  «*. 

54. 

^4  .+6  ^  l^2s+4^ 

75. 

^.'■+•+4  ^  ff^\ 

34. 

c^-f-c'. 

55. 

^6n+6  ^  ,.4«+6, 

76. 

«x+v+7  _^  ^x+,/+l^ 

35. 

-d^-^c^. 

56. 

^-^^^ 

77. 

^,«+*+c  _^  ^+5^ 

36. 

(-^)*-(- 

-^)- 

.57. 

g'^-^g^'- 

78. 

yn+n+8  _i_  ^y»>+8^ 

37. 

(-^)'-(- 

-kr 

.58. 

hr  ^  hr\ 

79. 

^+*+4  _i_  ^+2^ 

38. 

a'"+i  -^  a. 

59. 

t+i  ^  i-\ 

80. 

^2m+n+l  _^  ^m+n+1 

39. 

^-+«  ^  b. 

60. 

fji^n  _^  ^«-l 

81. 

^3r+2a+4  _:_  ^^8r+s+8 

40. 

c^"  H-  c". 

61. 

;,8r+2  ^  ^2.-. 

82. 

^4a+8*+2 '  _1_  ^  8  r+2*+c 

41. 

<^*^  -f-  G^^^ 

62. 

^4x+3^^4x-6 

83. 

^a+«*+7  c_^  w4 a+8*+3c^ 

42. 

/^-^^'. 

63. 

^«+n  ^  ^2n-l^ 

Division  of  One  Monomial  by  Another 

7.  It  may  happen  that  the  factors  of  a  given  divisor  may  be 
found  by  inspection  among  the  factors  of  the  dividend. 

In  this  case  the  quotient  is  by  definition  that  part  of  the  dividend 
remaining  after  striking  out  the  factors  of  the  dividend  which  are 
equal  to  those  of  the  given  divisor. 

Ex.1.    Divide  a:2/2  by  2/- 

We   may   write    xyz  -^  y  =  xzy  -^  y.     By    the    Commutative    Law    for 

Multiplication. 
^  xz.     Since  x  y  —  y  may  be  neglected. 

Ex.  2.    Divide  20  x*  l>y  5x. 

20  is  4  X  5,  a:*  is  x^  x  x. 
Hence,  20afi  -^  5x  =  (-ix^  x  5x)  -^  5x  =  4x^. 


DIVISION  llo 

Ex.  3.  If  we  are  required  to  divide  mn  by  a:,  then,  since  x  does  not 
*' appear"  as  a  factor  of  the  dividend  m?i,  we  may  indicate  the  quotient  by 

.  .  mn 

writing  mn  -^  x  =  — . 

X 

Ex.  4.    Divide  \2abh^  by  A:ahH, 

12  ahh^  ^  4abH  =  (12  -^  4)  x  (a  -f  «)  x  (6^  ^  h^-)  X  {c^  -f  d) 

a 
Ex.  5.    ISx^yz  ^  7x^yw  =  (18  -J-  7)  x  (x^  -i-  x"^)  x  {y  ^  y)  x  {z  ^  to) 

=  18  -i-  7  X  c  -^  w 

=  182 -^  /  w z=- — . 

Iw 
Let  the  student  check  these  examples  numerically. 

8.  The  division  of  one  integral  monomial  by  another  may  be 
indicated  by  writing  the  divisor  beneath  the  dividend,  separating 
the  two  by  a  horizontal  stroke  of  division. 

It  follows  that,  since  the  operation  of  division  introduces  neither 
the  symbol  of  addition  +,  nor  the  symbol  of  subtraction  — ,  that  the 
quotient  obtained  by  dividing  one  monomial  by  another  is  always  a 
monomial. 

The  quotient  obtained  by  dividing  one  monomial  by  another^  is  the 
quotient  of  their  numerical  coefficients  (considered  as  positive  or  nega- 
tive numbers)^  multiplied  by  the  quotient  of  their  literal  factors. 

Mental  Exercise  VIII.     2 

Perform  the  following  indicated  divisions  : 
Divisor)  Pi  vidend 
Quotient. 
\.2b^b\,  7.  -8ar)-16^'       13.  ^ab)--\2a''b\ 

2.  3c)12c^  8.  2a^)8^«.  14.  4cc?)8cU 

3.  r)d}26d\  9.  Sb^^)l5b\  15.  —3ga:)21ga:\ 

4.  4^)24^'.  10.  4c^)20c^  16.  -lla%)-22a%^ 
r>.  Gm)-30m\  11.  -9c?')-18^^  17.  12  b^k' )  36  bU:\ 

6.  -ln)2Hn\         12.  7/^')  U  h\  18.  dxyz) —  21  .tYz\ 


116  FIRST  COURSE  IN  ALGEBRA 

1 9.  1 5  a%  )30a»6V.  35.63  a?ys«  -^  7  ^z-y^r^. 

20.  -  13 acV)  52 aW.  ^^-  ^^  ^^^'^'^^  ^  ("  1^^^^> 

37.  —  45  m^nV  -^  9  w22wV. 

21.  14«W)^6«W,/.  33    ^^^,,,,  ^  .^.  ^,,, 

22.  16  a^fz)-ma:VzV.        39.  51  .^Vf'  ■^-  (-  H  ;^y;^*). 

23.  -  15  a%Y)loa'b'xy.        ^0.  (-  57  a W)  -  ("  19  a^/^V). 

41.  drlr-^ah, 

24.  -9^>'cV)-72^VyV. 

25.  42  rt"  ^  6  a=^.  43.  w^^w"  ^  ww. 

26.  39  ^'  -f-  13  h\  44.  ^r^y  ^  ;r^?^. 

27.  72  tf'A'  -^  12  a%\  ^.  a^'^^^n  ^  ^m^^ 

28.  56  h'^d'  -T-  8  b'^d^,  46.  ^*y  ^  -=-  ^>'y. 

29.  44  .ry  -^  (—  11  A-y).  47.  c^+^i^+i  -i-  cw;. 

30.  75  7nV  -^  (-  15  wV).  48.  ^"+V+^  -^  ^V. 

31.  (-  90  n^z')  H-  (-  10  nh"").  49.  <^'-+*/+2  ^  ^r+y^ 

32.  (-  96  gVr)  -^  (-  16  g'h*),  50.  a^m+s^sn+a  _^  ^2^2 

33.  (-  72  h'P)  ^  18  ^F.  51.  c»^+^(^^-^  -^  c^-^-^^^^+l 

34.  80  aVc'  -^  1 6  a^Z/^^^.  52.  a:*-+^Y"^'' '  -^  ^2«+y  .+3  c^ 

Division  of  a  Polynomial 'by  a  Monomial. 

9.  The  quotient  resulting  from  the  division  of  a  polynomial  by  a 
monomial  may  be  obtained  as  a  direct  application  of  the  Distributive 
Law  for  division.     That  is,  since 

(a  -\-  b  —  c)-r-d  =  a-T-d+b-^d— c-^df 

it  follows  that  we  may  divide  each  term  of  the  polpiomial  dividend 
by  the  monomial  divisor  and  write  the  algebraic  sum  of  the  resulting 
partial  quotients. 

Ex.  1.    Divide         15  a^b"^  -  10  a%^  +  5  ab^  by  562. 
(15  a^62  _  iQ  a2j3  _j.  5  ^65)  ^  5  52  =  15  ^4^,2  _^  5  j^2  _  xo  a%^  ^  5  Z;^  +  5  ah^  ^  5  // 

=  3a*  -  2a%  +  abK 

Check.     Let  a  =  3,  &  =  2. 
(4860  -  720  +  480)  -f  20  =  243  -36  +  2^ 
231  =  231. 


DIVISION  117 

Mental  Exercise  VIIL     3 
Perform  the  following  indicated  divisions: 
1.  2a)_6ab_±_8ac.  10.  3a^)Qa^h+  12a^c. 

I     2.  Sb)12bc+  186.r.  11.  ^^ . 

4  c^  —  8  c^d^ 
3.  Qd)S()da\i/  +  42 d.  12.  j^^^- — 


4.  5m)  5m  —  \Omn.  13. 

5.  7w)  14n^+  21nx.  14. 


4c^ 
4a6c  4-  10  abd 


2ab 

,„  ,^  ,^    32  7?^ V  -  40  ?w V 

6.  4c)8c7/  — 28fm  15.  ; . 

7.  8a?)24ir?/+  1657^.  16-  — -^i • 

8.  9.)27^.  +  45.v%.  17.  l^^^^-^f^V^ 

9.  6».)30^V+48;yV.         18.  l^m'^V -^^  >»'.'. 

"^^ ' —  4  w^.«r 

lOM^c  -  15 «2^c2  +  5 a^bc 


20. 
21. 
22. 


5  a%c 
da^bc^  +  12  aW;  +  24ff7>V 

3a6c 
21  a^b'cy'  -  7  r?^/;«c^  -  28  a^b^c^ 

7  «^/.2c=^ 
30  a^fz*  +  12  ^y^'  —1 8  ^^v/V 


-    6^y;2« 
25  m^nh^  —  30  ??z^;/^^'  —  45  m^n'^z^ 
^^'  -5mbih' 

28  T^sV  +  32  r«.s«M;'  —  44  rVw« 


24. 
25. 

26. 


4  r^sV 
40a^bd^  +  24  g'^^^'^^^  +  32a^<^«(^* 

8a2/^^8  .   * 

27  m'^nh^  —  18  ^^ "y^V  —  54  ??2  V^" 
-  9  m^n^z^ 


33^y;j^- 

Ua:Yz'- 

22  itYz^ 

-24a«' 

-  7  g'h'k* 

-  12  a'c'h' 

-48«Wi« 

-  2G  bVt'n' 

-  12  a^c'h' 

-  IS  b^h*n'' 

-  SdlW?i' 

118  FIRST  COURSE  IN  ALGEBRA 

27. 
28. 
29. 

-I3  6V*V 

31.  (f  a»^c  +  §  ff^'c  +  ^  t/^»c«)  H-  2  «^c. 

32.  *(f  a«Z;V  +  ia'^V  +  ^a^^V)  ^  Sa^b^c^ 

33.  (i  ^^^2  -  2  a-Vz  -  i  a^Y^  -  i  ^^^. 

34.  (^  m^n^w*  —  \^  m^n^w^  —  \^  m*nV)  -^  5  m^nV. 

35.  (I  a^bd'^  -  ^  a%H  -  i  a^bH")  -r-  (-  J  a%d). 

36.  [3(a  +  6)«  +  9(a  +  6)*  +  C(«  +  ^)^]  -^  3(a  +  b)\ 

37.  [5(.i-  +  yY  -  lh{x  +  ?/)'  -  S^{x  +  ^)*]  -^  5(  ar  +  y)\ 

38.  [4(6  -  cO'  +  20(/>  -  ay  +  16(6  -  df^^  ^  4(6  -  cQ'- 

39.  [6(^«  -  ^^  4-  18(^  -  lif  +  12(i7  -  /0«]  -  6(i/  -  /O. 

40.  \ar^  +  a^^^  +  «'"+*)  -^  a. 

41.  (6"+'  +  6-+«  +  6"+«)  -^  6«. 

42.  (c^ "+«  +  c^*  "+2  +  c^  '"*"0  ^  ^'"^  "• 

43.  (<^"+2  -  <?"+*  +  <^'*+*)  -7-  ^'^^^ 

44.  (;r»  "+*  —  .i^  "+»  —  ;r«  "^'^  -^  ^^  "+'. 

45.  (2^M- /"  +  //'* +  /0^2/''- 

46.  (;J^  "  +  ;;*''  +  -;°''  +  ;:^  ")  -4-  2;  ^. 

47.  («*'+"  +  «»^+-''  +  a^x+Si,  _|_  ^x+4y)  ^  ^x4tr^ 

48       f/"'+4»  7  2m+8«     I      /8w+2»  lAm+nX   _j_    1  m+n 

49.      (^7^'"+^  +  rt2"'+265  +  a2-+8^T)  ^  ^2n,^2_ 

Division  of  One  Polynomial  by  Another. 

10.  An  expressed  quotient  is  called  a  fraction ;  the  dividend  is 
called  the  numerator  and  the  divisor  the  denominator. 

11.  The  process  by  which  the  division  of  one  polynomial  by  an- 
other is  performed,  may  be  made  to  depend  upon  the  following 

Fundamental  Principle :  The  quotient  obtained  by  dividing  an 
integral  function  of  x  by  an  integral  function  of  x  which  is  of  degree 
not  higher  than  that  of  the  dividend  can  be  transformed  into  the  sum 
of  an  integral  function  of  x  (the  integral  quotient)  and  a  fraction  the 


DIVISION  119 

numerator  of  which  is  the  remainder  resulting  from  the  division^  and 
the  denominator  of  which  is  the  given  divisor. 

The  degree  of  the  integral  quotient  obtained  by  the  above  Division 
Transformation  is  equal  to  the  excess  of  the  degree  of  the  dividend 
over  that  of  the  divisor. 

The  degree  of  the  remainder  which  is  used  as  the  numerator  of 
the  fractional  part  of  the  transformed  function  is  accordingly  less 
than  the  degree  of  the  divisor  which  is  used  as  a  denominator. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  the  dividend  D  and  the  divisor  d  be  integral  functions  of  some  letter 
X,  the  degree,of  the  divisor  being  not  higher  than  that  of  the  dividend  with 
reference  to  x. 

Letting  Q  stand  for  the  integral  part  of  the  quotient  and  R  for  the 
remainder,  we  have 

We  may  construct  the  following  identity  : 

D-^d  =  D^d  +  Q-Q.    Since  Q-Q  =  0. 

=  Q  +  (J^  -7-  d  —  Q).     Commutative  and  Associative  Laws. 

=  Q  -}-(/)  _j-  (/_  Q)  xd-^d.     Since  x  d  ^  d  =  1. 

=  Q  -\-  (D  -^  d  X  d  —  Q  X  d)  -^  d.     Distributive  Law. 

=  Q  +  (D  -  Qd)  ^  d.     Since  -r- d  x  d  =  I. 

=  Q  -h  B  -^  d.     Since  D  —  Qd  is  by  definition  the  same  as  B. 

Division  of  one  Polynomial  by  Another. 

Development  of  the  Process 

12.  In  order  to  clearly  understand  the  process  of  division,  it  is 
well  to  obtain  the  product  of  two  given  integral  polynomials,  and 
then,  after  having  examined  carefully  the  manner  in  which  it  is  built 
up,  to  reverse  certain  of  the  steps  and  processes.  Then  starting 
with  the  reduced  product  as  a  dividend,  and  one  of  the  factors, 
say  the  multiplicand,  as  a  divisor,  wo  may  find  the  other,  the 
multiplier,  which  we  shall  now  call  the  quotient. 

13.  For  convenience,  we  shall  select  two  polynomials  in  which  no  powers 
are  missing,  and  shall  arrange  them  according  to  descending  powers  of  the 
same  letter,  say  a. 


120  FIRST  COURSE  IN  ALGEBRA 

Multiply  a*  +  3  a^fe  +  3  ah'^  +  6»  by  a^  4.  2  aft  +  h\ 

Form  I 

Multiplicand      .     .     .  a*  +  3  a-h  +  3  ah^     +6* Divisor. 

Multiplier     .     .     .     .  a^  +  2  a6  +  6^ Quotient. 

p      .  .        M  st  row  a^  +  3  a*6  +    3  a'ft'-^  +       a-l)^ 
Partial       \  ^^^^^  ^^^^^^        +  2 a*6  +    6  a^^^  4.    d  a^&s  +  2 aft* 

FnKlucts.      (3j.Jj.^^^,  +       q8Z>-^4-    3a-^^8  +  3a/;^  +  6^ 

Reduced  Product      .  a^  +  5  a*6  +  10  a^i^  _|_  10  a'-^f/^  +  5  af;*  +  />*  Dividend. 

14.  Observe  that  the  number  of  terms  in  each  horizontal  row  0/ 
partial  products  corresponds  to  the  number  of  terms  in  the  multipli- 
cand^ and  there  are  as  many  rows  as  there  are  separate  terms  in  the 
multiplier.  The  degree  of  the  first  term  of  each  row  with  reference 
to  the  letter  of  arrangement  is  higher  than  that  of  any  following 
term. 

15.  If  now  we  interchange  the  given  pol3naomials  and  use  the 

first  as  a  multiplier,  and  the  second  as  a  multiplicand,  we  shall 

obtain  the  same  reduced  product  and  the  same  partial  products  as 

before;  but  their  orders  of  arrangement  will  be  different,  as  in 

Form  II. 

Form  II 

Check  for  both  forms. 

Let  a  =  3,  ?*  =  2. 

Multiplicand     .     .  rt2  +  2a6  +6^ 25 

Multiplier    .     .     .  a^  +  3  ft^6  +  3  a/>^      -\-h^ J^5 

I  1st  row  rt5  4.  2  a%  +       a*P  3125 

2uil  row        +3  a*6  +    6  a^P-  +    3  a%^ 
3rd  row  +    ^aW -^    ^a%^-\-?,ah^ 

4th  row  +       a%^  +  2  ah^  +  h^ 

Reduced  Product         a^  ^o  a^h  +  10  a^W-  +  10  a%^  +  5  ah^  +  i^.     .     .  3125 

""o~ 

16.  Each  horizontal  row  of  partial  products  in  Form  II  corre- 
sponds to  an  oblique  or  diagonal  row  containing  the  same  terms  in 
Form  I,  and  each  row  in  Form  I  has  a  corresponding  oblique  or 
diagonal  row  in  Form  II. 

E.  g.  The  first  horizontal  row  in  Form  II,  a^  +  2rt*6  +  a%'^^  appears  as 
the  fii-st  diagonal  row  a^  +  2a*6  +  a%'^  in  Form  I  C§  13).     Also,  the  terms 


DIVISION  121 

in  the  first  horizontal  row  of  Form  I,  a^  +  3  a%  +  3  a%'^  +  a%^,  are  found  as 
the  first  terms  of  the  different  rows  of  Form  II;  that  is,  they  occur  in  the 
first  diagonal  row.     The  same  is  true  for  the  other  rows  of  partial  products. 

17.  The  arrangement  of  the  partial  products  for  a  given  example 
depends  simply  on  which  of  the  two  given  polynomials  is  chosen  as 
multiplicand,  and  which  as  multiplier. 

18.  In  either  Form  I  or  Form  II,  the  terms  of  any  particular  row 
are  obtained  by  multiplying  successively  the  terms  of  the  multipli- 
cand by  one  of  the  terms  of  the  multiplier. 

E.  g.  To  obtain  the  first  row  of  partial  products  use  as  a  multiplier  the  first 
tei'm  of  the  polynomial  chosen  as  multiplier;  for  the  second  row  use  the  second 
term  of  the  polynomial  multiplier;  for  the  third  row  the  third  term,  etc. 

19.  The  first  term  of  each  row  is  obtained  by  multiplying  the 
first  term  of  the  multiplicand  by  the  term  of  the  multiplier  corre- 
sponding to  the  number  of  the  row. 

E.g.  r    a^  is  obtained  by  multiplying  rt' by  6^;  (1) 

In  Form  I  ]  2  a*6  «  "  ""        a^  by  2  ah ;  (2) 

(.    a862  "  «  «         a'byi^.  (3) 

Similar  results  may  be  seen  to  be  true  by  examining  Form  II.  (§  15). 

20.  It  follows  that,  if  we  know  the  first  term  of  any  specified  row, 
we  may,  by  dividing  it  by  the  first  term  of  the  multiplicand,  obtain 
the  term  of  the  polynomial  multiplier  corresponding  in  number  to 
the  number  of  the  row. 

E.  g.    In  Form  I  (§  13)  we  have : 

From  the  first  row  of  partial  products a^  -^  a^  =  a^,        (1) 

which  is  the  first  term  of  the  polynomial  multiplier. 
From  the  second  row  of  partial  products  ...  2  a*6  -^  a^  =  2  ah,     (2) 

which  is  the  second  term  of  the  polynomial  multiplier. 
From  the  third  row  of  partial  products     .     .     .    a^b'^  -^  a^  =  b^,         (3) 
which  is  the  third  term  of  the  polynomial  multiplier. 

21.  From  the  arrangement  of  the  partial  products  in  columns, 
the  first  partial  product,  a^,  is  the  first  term  a^  of  the  reduced  pro- 
duct in  either  Form  I  or  Form  II.     (§§  18,  15). 

22.  If  now,  returning  to  Form  I,  we  divide  a^  by  a^,  we  obtain  as 
a  quotient  the  first  term  a^  of  the  polynomial  multiplier. 

That  is,  a^  -^  a»  =  al     (See  (1)  §  20.) 


122  FIRST  COURSE   IN  ALGEBRA 

23.  If  we  multiply  the  terms  of  the  multiplicand  successively  hy 
this  multiplier  J  a^,  we  obtain  the  first  horizontal  row  of  partial  prod- 
ucts as  in  Form  I  (§  13).  By  subtracting  the  terms  of  this  first  row 
from  the  reduced  product,  we  have  as  below  : 


Step  (i. 


■Reduced  product,  a^  +  5  a^h  +  10  a^h"^  +  10  a%^  +  bah^  +  h^ 

First  row  of  partial 

proilucts  ffS  +  3a^6+    Za^"^-^      a%^ 

Fii-st  partial  remainder,  2a*b+    7a%'^-{-   9  a%^ -{- 5  ab*  +  b^. 


24.  The  result  of  the  subtraction,  2  «*6  +  7  a^O'  +  9  a%^  +  5ab*  + 
b\  we  shall  call  the  first  partial  remainder.  From  the  nature 
of  the  case  the  first  term,  2ci%  is  the  same  as  the  first  term  of 
the  second  row  of  partial  products  in  Form  I  (§  13).  It  is  also  the 
sum  of  all  of  the  partial  products  in  Form  I,  except  those  in  the 
first  row  which  were  subtracted. 

25.  Dividing  the  first  term  2  a*b  of  the  first  partial  remainder  hy 
a^  we  obtain  2  a^b  H-  «*  =  2  ah.     (See  (2)  §  20.) 

The  term  2  ab  thus  obtained  is  equal  to  the  second  term  of  the 
original  multiplier.  Referring  to  Form  I  (§  13)  it  may  be  seen 
that  this  step  in  the  process  consists  in  dividing  the  term  of  highest 
degree  with  reference  to  the  letter  of  arrangement  (that  is,  the  first 
term  of  the  second  row  of  partial  products  in  Form  I)  by  the  first 
terra  of  the  multiplicand  above. 

26.  By  multiplying  the  multiplicand,  as  in  Form  I  (§  13),  by 
2  a6  as  a  multiplier,  we  obtain  the  second  row  of  partial  products 
2  «*/>  +  6  rt»^2  +  6  a'^'  +  2  ab\ 

27.  Subtracting  these  terms  from  tlie  first  partial  remainder 
obtained  above  in  §  23  (i.),  we  have 

f  Fi rst  partial  remainder,  2  aSh  +  7  a^"^  +  9  a^fts  +  5  a54  +  56^ 

Step  (ii.)  -  Second  row  of  partial  products,  2  a^6  +  6  a%'^  +  6  a^lfi  +  2  ab\ 

VSecond  partial  remainder,  a^^^  +  3  a^"^  +  3  a6*  +  b^- 

28.  The  first  terma*6^  of  the  second  partial  remainder  is  the 
same  as  the  first  term  of  the  third  row  of  partial  products  in 
Form  I  (§13). 

29.  As  we  proceed  in  our  work  any  particular  remainder  will. 


DIVISION  123 

from  the  nature  of  the  case,  be  the  sum  of  the  rows  of  partial  pro- 
ducts below  the  last  one  subtracted.     (See  Form  I,  §  13.) 

30.  Dividing,  as  before,  the  first  term  a^iP"  by  the  first  term  of  the 
multiplicand  a^  we  have  as  in  (3)  §  20, 

which  is  the  third  term  of  the  original  polynomial  multiplier. 

31.  As  before,  we  use  this  as  a  multiplier  with  the  entire  multi- 
plicand, and  obtain  the  third  row  of  partial  products  in  Form  I 
(§  13),  a^'U'  -f  3a'^«  +  3  ah''  +  h\ 

32.  Subtracting  tliese  terms  from  the  second  partial  remainder^  we 
find  as  below,  that  the  third  partial  remainder  vanishes,  that  is,  it 
is  zero  : 

(  Second  partial  reiriainder,  a^V^  -f  3  a%^  +  3  «i*  +  h^ 

Step  (iii.)  \  Third  row  of  partial  products,    a^lP-  +  3  a^^^  +  3  uh'^  +  h^ 
(  0 

33.  The  process  stops  here,  as  we  should  naturally  expect,  since 
we  have  recovered  all  of  the  terms  of  the  original  polynomial  multi- 
plier a^,  +  2  ah  and  b\     (See  Form  I,  §  13.) 

34.  Summary.  By  this  process  we  have  succeeded  in  finding 
the  original  polynomial  multiplier,  when  the  reduced  product  and 
the  original  multiplicand  were  given.  The  process  would  have  been 
exactly  the  same  if  we  had  started  with  the  reduced  product  and 
the  original  multiplier,  a^  +  2ab  +  b^,  to  find  the  original  multipli- 
cand a^  +  3  a*^6  +  3  ab^  +  b^.  In  that  case,  however,  there  would 
have  been  this  exception,  that  the  successive  rows  of  partial  products 
would  have  corresponded  to  the  rows  of  partial  products  in  Form  II 
(§  1 5),  where  the  polynomial  to  be  found  is  used  as  a  multiplier. 

35.  We  have  thus  developed  a  method  for  finding  the  original 
multiplier  when  the  multiplicand  and  the  reduced  product  are 
given.  The  method  holds  good  also  for  finding  the  original  mul- 
tiplicand when  the  multiplier  and  the  reduced  product  are  given. 

36.  It  will  be  seen  from  the  discussion  above  that  the  reduced 
product  takes  the  place,  in  our  problem  of  division,  of  the  dividend, 
the  original  multiplicand  of  the  divisor,  and  the  original  multiplier 
of  the  quotient. 


124  FIRST  COURSE  IN  ALGEBRA 

Correspondence  between  Examples 


IN 


DIVISION  MULTIPLICATION 


DIVIDEND 


DIVISOR MULTIPLICAND 


QUOTIENT ....  MULTIPLIER 
REDUCED  PRODUCT 


37.  The  rows  of  partial  products  in  the  process  of  multiplication 
correspond  to  those  appearing  in  the  process  of  division. 

38.  The  Division  Transformation.  Whenever  division  is 
possible,  we  may  carry  out  the  different  steps  of  the  process  as 
follows  : 

First  arrange  the  terms  of  both  dividend  and  divisor  accord- 
ing to  ascending  oi'  descending  powers  of  some  letter. 

Place  the  divisor,  for  convenience^  at  the  right  of  the  dividend. 
Since  the  different  terms  of  the  quotient  are  to  be  used  as  multipliers 
during  the  process,  it  will  be  convenient  to  write  the  quotient,  term  by 
term,  immediately  below  the  divisor. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the  divi- 
sor, and  ivrite  tJie  result  as  the  first  term  of  the  quotient. 

Multiply  the  whole  divisor  by  this  first  term  of  tlie  quotient,  and 
write  the  ?'esulti?ig  partial  products  under  the  dividend. 

Subtract  from  the  dividend  the  polynomial  composed  of  the  partial 
products,  and  bring  down  the  result  as  the  first  partial  remainder. 

Divitle  the  first  term  of  the  first  partial  remainder  by  the  first 
term  of  the  divisor  as  before,  and  write  the  result  as  the  second  term 
of  the  quotient. 

31ultiplf/  the  whole  divisor  by  the  second  term  of  the  quotient,  sub- 
tract the  resulting  product  from  the  first  partial  remainder,  and 
write  the  result  as  a  second  partial  remainder. 

Repeat  the  operations  above  either  until  the  remainder  is  0,  or 
until  as  many  terms  of  the  quotient  are  found  as  are  desired.  In  the 
latter  case,  add  algebraically  a  fraction  having  for  numerator  the 
remainder  at  this  stage,  and  for  denominator  the  divisor. 


DIVISION 


125 


Arrangement  of  the  Work  • 

39.   A  convenient  arrangement  of  the  different  steps  of  the  process 


is  shown  below  : 

DIVIDBND 

Multiplying  divi-    a^+5a*b+10a%-^+10a%^+5a¥+h^ 


801-  bv  a^ 


««f  3(1*6  +  3a%'^+     a%» 


1st  partial  remainder  .     2a^b-\- 
Miilt'g  divisor  by  2ab  .     2a%  + 
2nd  partial  remainder  .     .     . 
Multiplying  divisor  by  b'^  .     . 


7a862+ 


9a%^+bab^+b^ 
6a268+2«6* 


a862+ 


3a%»+'3ab*  +  b^ 
3a%»-{-3ab*-\-b^ 


a»+Sa%-h3a¥-^b^ 


a^+2  ab  +  b^ 

QUOTIBNT 

Check.     Let 
a  =  3,      &  =  2. 

3125)125 
Quot't  sh'd  be  25 
Quotient  is    .     25 

~0 

"We  have  carried  out  the  process  above » on  the  assumption  that 
the  degree  of  the  dividend,  with  reference  to  the  letter  of  arrange- 
ment, was  at  least  as  high  as  that  of  the  divisor,  —  that  is,  that 
division  was  possible.  It  is  the  exception  rather  than  the  rule  that 
we  find  an  integral  quotient  when  dividing  one  integral  polynomial 
by  another.  We  therefore  apply  the  Principle  of  No  Exception, 
and  assert  that  division  may  be  performed,  if  indeed  it  can  be  begun 
at  all,  by  the  steps  of  the  process  above. 

40.   In  the  division  transformation  two  cases  may  arise  : 
First,  it  may  be  possible,  or  second,  it  may  be  impossible,  to  find 
as  a  quotient  an  integral  function  of  x. 

In  the  first  case  the  division,  if  carried  out,  is  said  to  be  exact 
and  there  is  no  remainder.  It  may  be  shown  that  when  division  is 
exact,  the  form  of  the  quotient  will  be  the  same  whether  the  division 
be  carried  out  with  both  dividend  and  divisor  arranged  according  to 
descending  or  ascending  powers  of  some  specified  letter. 

Ex.  1.   Divide  a;2  +  lOx  +  21  by  a;  +  3. 

The  process  is  shown  below,  at  the  left  with  the  dividend  and  divisor 
arranged  according  to  descending  powers,  and  at  the  right  with  both  arranged 
according  to  ascending  powers. 

Descending  Powers.  Ascending  Powers. 


a:2  +  lOx  +  21 
X24.    Sx 


a: +  3 


x  +  7 


21  + 
21  + 


10a;  +  a;2 

7x 


3  + a: 


7+x 


7a:  +  21 
7  a: +  21 


3  a;  +  x'^ 
3a:  +  a:2 


The  student  should  check  the  example  numerically. 


126  FIRST   COURSE  IN  ALGEBRA 

4i.  The  form  of  the  quotient  obtained  when  both  dividend  and 
divisor  are  arranged  according  to  ascending  powers  of  some  letter  of 
arrangement,  is  not  the  same  as  that  of  the  quotient  obtained  when 
the  terms  are  arranged  according  to  descending  powers. 

Ex.  2.   Divide  8  x^  +  17  x  +  1 4  by  a:  +  2. 

The  process  is  shown  below,  at  the  left  with  l)otb  dividend  and  divisor 
arranged  according  to  descending  powers  ofx,  and  at  the  right  with  both 
arranged  according  to  ascending  powers  of  x.  Let  the  student  check  each 
result  numerically. 

Descending  Powers.  Ascending  Powers. 


8x2+i7a:-f-  14 

x  +  2 

8x2+  lOa; 

«^+^'+x!:2 

14+  17x  +  8rc2 

2  +  x 

14+    7x 

^  +  ^^  +  2  +  x 

x+14  10x  +  8x2 

x+    2  10x+  5x2 

12  3^ 

42.  When  the  operation  of  division  is  performed  with  both  divi- 
dend and  divisor  arranged  according  to  descending  powers  of  some 
letter,  it  will  happen  either  that  division  will  be  exact,  or  that 
we  shall  arrive  sooner  or  later  at  a  remainder  which  is  of  lower 
degree  with  reference  to  the  letter  of  arrangement  than  the  divisor. 
Here,  for  the  present,  the  operation  of  division  terminates. 

If,  however,  division  be  carried  out  with  both  dividend  and 
divisor  arranged  according  to  ascending  powers  of  some  letter,  it 
will  happen  that  when  division  is  not  exact  the  degrees  of  the 
remainders  will  be  successively  higher  and  higher,  and  an  un- 
limited number  of  terms  may  be  obtained  in  the  quotient. 

Ex.  3.    Divide  3  x  +  4  x^  by  1  +  x  +  x^. 


3x  +  4x2 

1   +  X  +  X2 

3x  +  3x2  +  3x8 

3x  +  x2-4x3+ 

X2  -  3  X8 

X2+       x3 

+     x^ 

-4X3-       2-4 

-4x3-4x4-4x5 

3x4  +  4x6 
etc. 

DIVISION 


127 


The  quotient  obtained  in  the  example  above  may  be  checked  at  any  stage 
of  the  process  by  adding  to  the  quotient  at  that  stage  a  fraction  whose  nu- 
merator is  the  corresponding  remainder  and  whose  denominator  is  the 
divisor,  and  then  making  numerical  substitutions. 

43.  We  may  continue  the  operation  of  divisjon  as  long  as  the  first 
term  of  the  arranged  partial  remainder  is  divisible  by  the  first  term 
of  the  divisor. 

By  bringing  down  the  terms  of  the  dividend  in  the  successive 
partial  remainders  only  as  they  are  actually  needed,  we  may  save 
labor  when  carrying  out  the  process. 

44.  Detached  Coefficients  in  Division.  Detached  coeffi- 
cients may  be  used  in  division  as  well  as  in  addition  and  mul- 
tiplication. 

To  illustrate  the  use  of  detached  coefficients  in  division,  the 
following  example  is  performed  first  in  the  ordinary  way  and  then 
again  by  using  detached  coefficients.* 


Ex.4. 
Ox'-lTarS-f-Slx*- 

_44a;8  +  r)5x2-40x«+14 
-18x8  +  21x2 

2x4-3x8  +  5x2_6x  +  7 
3x2 -4x  _,_2 

-  8x5  +  16a;*- 

-  8x6  +  12x4- 

-26x8  + 34x2 -40x 
-20x8  +  24x2-28x 

Check.     Let  x  =  2. 
138)23 

4x*- 
•4x*- 

-6x8+  i()a;2_  12X+14 
-    6x«+  10x2-  12x+  14 

Quotient  should  be   6 
Quotient  is                6 

0 

Using  detached  coefficients : 


6 

-17 
-    9 

+  31 
+  15 

-44 

-18 

+  55 

+  21 

-  40     +  14 

2      -  3      +  5 

-6     +7 

6 

3      -  4      +2 
3  x2  -  4  X   +2 

-  8 

-  8 

+  16 
+  12 

-26 
-20 

+  34 
+  24 

-40 

-28 

4 
4 

-  6 

-  6 

+  10 
+  10 

-12     +14 
-  12     +  14 

45.  Numerical  Checks  in  Division.  Since  zero  cannot  be 
used  as  a  divisor,  it  follows  that  care  must  be  taken  when  employ- 
ing numerical  checks  in  division  to  avoid  giving  to  the  letters  such 
values  as  would  cause  the  divisor  to  become  zero. 

E.  g.  When  checking  the  quotient  obtained  by  dividing  x2  —  7  x  +  1 2 
by  X  —  3,  we  must  avoid  giving  the  value  3  to  x,  for  in  this  case  the 
'li visor  X  —  3  would  represent  the  value  zero. 


128  FIRST  COURSE   IN   ALGEBRA 

Exercise  VIII.    4 
When    performing    the    following  divisions,   check    all    results 
numerically.     Divide : 

1.  a^+  6a  +  8bya  +  2. 

2.  Qb^  +  56  — 6  hy  36- 2. 

3.  5c«  +  c-6by  5c+  6. 

4.  21c?^  + 386?+ 16by  7(3?+ 8. 

5.  1  +  2c  +  2c2  +  c*by  1  +  c. 

6.  6a'^  +  8a  +  28  by  3a  +  7. 

7.  6a*^+ 13a6  + 66^by  2a  +  36. 

8.  8a;'*  — 22a^+  15/by  2ic  — 3^^. 

9.  a«  +  b*  by  a^  +  b\ 

10.  w*  +  2  7^*  +  3  m^n  +  4  mn^  by  ??2  +  7i. 

11.  P  +  Ar^w  —  km"^  —  w*  by  /:  —  m. 

12.  a*  -  2a»6  +  2a6*  -  6*  by  a''  -  b^. 

13.  1635*— 1  by  2«— 1. 

14.  32  TW*^  +  1  by  2  w  +  1. 

15.  3a«-24bya^-2. 

16.  Q(P-(P-12d+  4.hyS€P+  4:d-\. 

17.  5*  +  ;:2««'2  +  m;*  by  z^  +  zw  +  w\ 

18.  ^*  -  ^  -  g2  _^  ^  by  ^'^  +  g  +  1. 

19.  ^*  -  6^«2  +  9AV  -  4«*  by  F  +  3^5;  -  2;^^ 

20.  c*  —  c»  —  Sc^  +  10c  —  10  by  c*  +  2c  -  2. 

21.  15a*  — a  +  8a'—  1  —  19a»  by  5a' -  .3a  —  1. 

22.  6/  —  13  icy«  +  13  £cy  -  13  x^y  —  5  a;*  by  2if  —  ?*xy  —  ^. 

23.  c^-6c*+  16c«  — 25c'+  13c  +  5byc»  — 4c'+  3c+  1. 

24.  a;*  —  4a^y  +  6ar*/  —  4a;/  +  ?/*  by  a;'  —  2 a;?/  +  /. 

25.  s^+  35*  — 20s»  — 60s'+64s  +  192bys'  +  9s'+  26.9  +  24. 

26.  3a;^  — 8  a;*  — 5a;»  +  26a;'-28a;+24bya;8-2a;'  — 4a;+8. 

27.  a^  +  10a«6'  +  6^  +  10a'6»  i-  5a*6  +  5a6*  by  a'  +  2a6+  6'. 

28.  3a;^+  7  a;*?/—  lla;y  —  11  a;'/ +  6a;?/*  —  18/ by  a;  +  3?/. 

29.  1  +  2  a'  —  7  a*  —  16  a**  by  1  +  2  a  +  3  a'  +  4  a^ 

30.  -y^  -  2  v«  +  1  by  ^^'  -  2  ?;  —  1. 

'  31.  3a;«  +  43a;' -  6a;»  —  30a;  +  80  — 32a;*  +  20a;^bya;+  8. 

32.  1 5  a*  +  1 6  a^  +  8  a*  —  9  a»  —  7  a'  +  1 9  a  —  42  by  5  a'  +  2  a  —  7. 

33.  c'  —  6  cV*  +  14  &d^  —  12  c*c?»  by  c^  -  2  c'^d^. 

34.  a;'  +  ?/'  +  c'  -  3  a;^^  by  a;  +  ?/  +  2;. 


DIVISION  129 

46.   The  ordinary  process  for  "  long  division  "  may  also  be  em- 
ployed when  the  coefficients  are  fractional. 
Ex.  1 .    Divide  ^ x^  +  -J^ xhj  +  lif  by  ^^x  +  \ y. 


l^  +\y 


ix^-^xy  +  ^y''. 


^xhj-lxy"^ 


^xy^  +  ly\ 
Let  the  student  check  this  example  numerically. 

Exercise  VIII.     5 
Divide  : 
1.  ^^*  -  ^^^y  +  ],xf  +  ^/  by  i./-  +  \y. 

2.  § b'' - f^e h^''  +  ni b' - m b'  +  m h' -^hy^i^ -^. 

3.  IP  -  H^'  +  m\k'  -  -^^l"  +  4^^^:*  by  ^F  -  Jy(:  +  }. 

4.  «^  -  -^rt*  +  10a«  -  30a'  +  90a  —  27  by  81a  —  27. 

6.  w'  — ^  »^«  +  TfVw^H^»^*-M^*-  V^^+  ^iw-lbylw'^-iwi  +  l. 

7.  a^  -  §Ja*  +  ^1^5  a'  +  Hi  «'  -  iM  ^^  +  f  by  §a'  -  ia  +  f 

8.  ^:«'-  |-|a;«  +  1^^  -  ^-x!"  +  f|a;«  -  4u;2by  f  x-«  -  f^  +  f. 

9.  «"•+'  +  a'"^  +  a^;"»  +  If'^''  by  a"*  +  l^. 

10.  2  a*'"  —  6  a^'^lr  +  6  a'"^'"'  —  2  //'"  by  2  a"*  —  2  IT. 

47.  The  Remainder  Theorem.  When  a  rational  integral 
expression  containiny  one  unknown^  Xj  arranged  according  to  descend- 
ing powers  of  x  is  divided  by  x  —  a,  the  remainder  may  be  obtained 
hy  substituting  a  for  x  in  the  original  expression. 

Let  .an  expression  or  function  of  x,  arranf,'ed  according  to  descending 
powers  of  a:,  be  represented  by  f{x).  When  f{pc)  is  divided  by  x  —  a,  denote 
the  quotient  by  Q  and  the  remainder  by  R. 

Then,  from  the  identical  relation  of  division,  we  have 

Kx)  =  {x-a)Q  +  B, 
Since  no  restriction  has  been  placed  upon  the  value  of  x  we  may  assign 
to  it  any  value  we  please,  such  as  a.     Representing  the  result  of  substitut- 
ing a  for  x  wherever  x  Jippears  in  the  identity  aVjove,  we  may  write 
/(a)  =  {a-  a)  Q  +  A'. 
9 


130  FIRST  COURSE  IN  ALGEBRA 

Since  (a  —  a)  is  0,  the  product  of  (a  —  a)  multiplied  by  the  quotient  Q 
vanishes,  and  we  have  /(a)  =  R. 

It  follows  that  we  may  obtain  the  remainder  after  division  by  x  —  a, 
by  writing  a  in  place  of  x  in  the  original  function. 

48.  The  remainder  obtained  by  dividing  an  expression  such  as 
Ax^  +  By?  +  Cotp-  +  Bx  +  ^^  by  a  binomial  divisor  x  —  a,  may  be 
obtained  by  replacing  x  in  the  original  expression  by  a.  We  may 
show  this  by  actually  carrying  out  the  process  of  division  as 
below  : 

Ax^—Aa:^ |via5»-|-(^a4-B)aB*+(^rt2+Ba+C)x 

■\-{^Aa-\-B)ofi-{Aa^B)ax'^ 

-^{Aa^^-Ba^C)x'i-{Aa^^Ba^C)ax 

-\-{Aa>-\-Ba'-{-Ca-{-I))x^E 
-h(Aa«+Bai+Ca+D)x-(Aa*-{-Ba»+Ca^+Da) 

Remainder 

It  may  be  seen  that  wherever  a  appears  in  the  remainder  Aa*  +  Ba*  + 
Ca^  +  Da -\-E,x  is  found  in  the  given  dividend  Ax*  +  Bx^  +  Cx^  +  Dx-{-E. 

49.  Synthetic  Division.  If  we  write  only  the  coefficients  of 
the  dividend  and  place  the  second  term  of  the  binomial  divisor 
with  its  sign  changed,  that  is,  —(—«)  =  +  a,  at  the  right  as 
below, 

-{■A     +  B    +  C    +  D    -\-  E)a 
we  may  obtain  the  coefficients  shown  in  heavy-lace  type  in  §  48 
by  the  following  process,  known  as  Synthetic  Division : 

+A  +  B  +   C  -\-D  +E  )  +  a 

+  Aa  +  {Aa^+Ba)       ■\-{Aa^  +  Ba'^+ Ca)    +  {Aa*  + Ba^+ 00^  + Da) 

+A+(Aa-\-B)  +(^a»  +  Ba+C)+  (Aa^ -{- Ba^-\-Ca+I)),-\-(  Aa*-\-Ba,^+Ca'^-\-I>a-]-E) 

Remainder. 

50.  Write  under  the  long  line  the  first  coefficient  -\-  A  oi  the 
original  expression,  then  multiply  it  by  a,  and  write  the  product 
Aa  diagonally  above  the  line  in  the  second  place,  that  is,  under  B. 
Then,  adding  Aa  and  B^  we  obtain  the  heavy-face  coefficient 
Aa  +  B,  which  is  written  immediately  underneath. 


DIVISION  131 

Continue  this  process,  that  is,  multiply  this  last  expression  Aa  +  B 
by  a ;  write  the  result,  Aa^  -\-  Ba,  above  the  line  under  G  in  the 
third  place.  Then  we  have,  adding  Aa^  ■{■  Ba  to  (7,  the  second 
heavy-face  coefficient  Aa^  +  Ba  +  C 

Continuing  this  process,  we  obtain  as  the  last  sum  the  remainder 
Aa*^  +  Ba^  +  Ca'^  +  I>a  +  E. 

Ex.  1.    Divide  4x^  +  3  x*  -  2  x^  +  x^  -  x  +  5  hy  x  -  2. 
Writing  the  coefficients  only,  with  +  2  in  the  divisor's  place,  we  may 
proceed  as  follows  : 

Coefficients  of  Dividend.  Modified  Synthetic  Divisor. 

4+    3_    2+    1  -    1   +      5  )  +2 
^  +  ^  +22  +  40  +  82  +162 
^4+11+20  +  41 +81, +  167 

^~-— — r TTTTr-r^        Remainder. 

Coemcients  of  Quotient. 

First  bring  down  the  4  underneath  the  line.  Multiplying  the  4  hy  2  we 
obtain  8  ;  adding  3,  11  ;  multiplying  by  2,  22;  combining  with  —  2,  +  20  ; 
multiplying  by  2,  +  40  ;  adding  1,  41 ;  multiplying  by  2,  82  ;  combining  with 
—  1,  81 ;  multiplying  by  2,  162,  which  when  combined  with  5  gives  the 
remainder  sought,  167. 

Using  as  coefficients  the  numbers  below  the  broken  line,  mn,  with  the 
exception  of  the  last,  we  may  construct  the  q\iotient  by  writing  in  the  literal 
factors.  Since  the  first  term  of  the  dividend  4a;^  divided  by  the  first  term 
of  the  divisor  x,  produces  the  quotient  4  x\  we  begin  in  the  first  term  with 
the  highest  power  x*. 

The  result  thus  obtained  is     4  x*  +  1 1  xS  +  20  a;^  +  41  a:  +  81  +  — -^  • 

61.  It  will  be  noticed  that  in  carrying  out  this  process  we  work 
back  and  forth  across  the  horizontal  line  mn  in  the  directions  in- 
dicated by  the  arrows  in  the  accompanying  figure.  We  begin  at 
4  and  arrive  finally  at  the  remainder  167. 

4  167 

''  52.  In  case  any  powers  of  the  pol3Tiomial  dividend  are  lacking, 
their  places  must  be  indicated,  when  this  process  is  applied,  by 
writing  in  terms  with  zero  coefficients,  as  in  the  following  example  : 


132  FIRST  COURSE  IN  ALGEBRA 

Ex.  2.    Divide  5a;*  +  3x2  -  2  by  a  -  3. 
This  expression  is  equivalent  to 

5x4  +  0a:3  +  3u;2  +  0a:-2. 
The  work  may  be  arranged  as  follows  : 

5+0+3+      0  -      2)  +  3 
_  +  15  +  45  +  144_  +  432 
5+ 15  +  48  + 144, +  430 

Remainder. 
The  highest  power  of  ar  in  the  quotient  is  x^,  since  5  x*  divided  by  x  is  5  x^. 

430 


We  have  as  a  result     5  x^  +  15  x^  +  48  x  +  144  + 


x-3 


53.  In  order  to  divide  an  expression  arranged  according  to  de- 
scending powers  of  a;  by  ic  +  a,  that  is,  by  ic  —  (—  a),  the  same 
process  is  employed,  except  that  —  a  is  used  in  the  same  way  as 
+  a  was  used  when  the  divisor  was  x  —  a. 

Ex.  3.  Find  the  remainder  when  4  x*  —  3  x*  —  5x2  —  x  +  10  is  clivided 
by  X  +  2. 

4-    3-    5-    1   +  10)-2 

-    8  +  22-34+70 
4-11  +  17-35, +  80 

Remainder. 

54.  If  when  a  function  of  x^  f{x)^  is  divided  hy  x  —  a  the  division 
is  exact,  that  is,  if  there  be  no  remainder,  the  identity /(a?)  =  (x  —  a)Q 
+  B  reduces  to /(a;)  =  (x  —  a)Q.  By  substituting  a  for  x  this  last 
identity  reduces  \xif{a)  =  0. 

Hence  we  have  the  following 

Factor  Theorem :  If,  when  a  is  substituted  for  x  in  an  expression 
arranged  according  to  descending  powers  of  x^  the  expression  becomes 
0,  then  X  —  a  will  be  an  exact  divisor  of  the  expression. 

Exercise  VIII.    6 
Find  the  quotient  and  remainder  when 

1.  ar'  +  5  a;  +  7  is  divided  by  a^  +  3. 

2.  ?/^  +  7  7/  +  11  is  divided  by  ?/  +  4. 

3.  z^—llz-\-  39  is  divided  by  ;j  +  3. 

4.  w'2  —  48  w  +  97  is  divided  by  ?r  +  10. 

5.  a*  +  11  a^—  19a  +  117  is  divided  by  a  +  4. 


DIVISION  133 

6.  ^»  -  19  6^  +  18  6  -  17  is  divided  by  6  -  5. 

7.  c»  -  14  c^  +  28  c  —  42  is  divided  by  c  —  G. 

8.  d^  -  bd""  +  6d^  -  lOd  +  1  is  divided  by  «^  +  1. 

9.  k^  +  2F  +  3F  +  4^  +  5  is  divided  by  /;  -  5. 

10.  3  a^  —  5  aj*  +  6  i«'  —  2  ic^  +  a;  —  1  is  divided  by  a;  —  2. 

Employing  the  Remainder  Theorem,  find  the  numerical  value  of 
the  following  expressions  when  x  is  given  the  value  indicated  : 

11.  x^  +  ?>x''  +  4a;  +  3,  when  a;  =  1. 

12.  2a^  +  a;^  +  7  a  +  4,  when  a;  =  1. 

13.  a^  4-  2  a;2  +  3  a;  +  5,  when  a;  =  2. 

14.  a;*  +  2  a;^  +  5  a;  +  7,  when  a;  =  2. 

15.  a;^  -  8  a;'*  +  17  a;  —  10,  when  a;  =  1. 

16.  a:*  +  i)x^  +  3aj  +  1,  when  a;  =  2. 

17.  a^  +  9  a;^  —  7  a;  +  5,  when  a;  =  1. 

18.  a^  —  7  a;2  +  13  a;  —  6,  when  a;  =  2. 

19.  3  a;»  +  2a;2  +  a;  +  5,  when  a;  =  2. 

20.  2  a.-*  —  5  a;''  +  4  a;  +  7,  when  a;  =  3. 

21.  3a;»  +  a;'  —  7 a;  +  9,  when  aj  =  1. 

22.  4a."«  -  9ar*  -  a;  H-  6,  when  a;  =  1. 

23.  2x^  +  x^  +  3  ^  +  2,  when  a;  =  2. 

24.  4a;«  —  3  a;2  +  2  a;  +  5,  when  aj  =  3. 

25.  3  x-*  +  7  a;2  —  4  aj  +  3,  when  a;  =  —  3. 

26.  4a.-*  —  5  a;^  +  3  a;  —  4,  when  a;  =  —  2. 

27.  2a.^  —  9  a;'  +  4  a;  +  5,  when  a;  =  5. 

28.  3  a;*  +  11  a;*^  —  7  a;  +  5,  when  a;  =  -  1. 

29.  2x'»  +  11  a;2  -  3  a;  +  10,  when  a;  =  -  5. 

30.  6  a.-'  -  3  a;2  +  7  a;  +  4,  when  a;  =  -  2. 

31.  11  Q^  —  9  a;2  +  7  a;  +  1,  when  a;  =  -  1. 

32.  a;^  +  2a.'»  +  3  a.-^  +  a;  +  1,  when  a;  =  1. 

33.  a;*  —  a;*  —  7  a;^  +  a;  —  6,  when  a;  =  1. 

34.  a:*  —  3  a^  +  a;2  +  2  a;  +  5,  when  a;  =  3. 

35.  a^  +  2  aj^  +  7,  wlien  a;  =  2. 

36.  a^  +  1,  when  a;  =  3. 

37.  a^  +  4  a;2  +  12,  when  a;  =  4. 

38.  Q^  +  ir)x+  3,  when  a;  =  5. 

39.  2  a;'  +  7  a;  +  20,  when  a;  =  2. 


134  FIRST  COURSE  IN  ALGEBRA 

40.  2  a^  +  5  ar»  —  33  ic  +  12,  when  x  =  3, 

41.  3s(^-Gx^+5x+  21,  when  x  =  2. 

42.  2  ic*  +  5  a'  4-  2  a  +  1,  when  x  =  I, 

43.  3  ic*  +  5  a^  +  4  a;  +  1  when  x  =  —  h 

Applications  of  the  Remainder  Theorem. 

55.   (i.)  TTie  binomial  difference  af"  —  y"'  is  always  divisible  with- 
out remainder  by  the  difference  x  —  y. 

For,  substituting  y  for  a;,  we  have  'f  —  y'"  =  0. 

Hence  by  the  Remainder  Theorem  the  division  is  exact. 

(ii.)  TTie  binomial  sum  af^  +  y'^  is  never  divisible  without  remain 
der  by  the  difference  x  —  y. 

For,  substituting  y  for  x,  we  have  f  +  y"^  =  2f^0. 

Hence  by  the  Remainder  Theorem  the  division  is  not  exact. 

(iii.)  The  binomial  difference  x""  —  y'"  ism-  is  not  divisible  without 
remainder  by  the  sum  x  +  y  according  as  m  is  even  m-  odd. 

For,  substituting  —  ^^  for  a^  we  have  {—yT  —  y^  (X)- 

Examining  this  "  remainder  "  for  both  even  and  odd  values  of  m, 
we  draw  the  following  conclusions  : 

If  m  be  even,  then,  since  all  even  powers  of  negative  numbers 
are  positive  numbers,  (— y)"*  —  y"  becomes  y^  —  y^  =  0. 

By  the  Remainder  Theorem  the  division  in  this  case  is  exact. 

If  m  be  odd,  then,  since  all  odd  powers  of  negative  numbers  are 
negative  numbers,  (—  y)"*  —  y"  becomes  — y^  —  y"'=.  —  2y^^0. 

By  the  Remainder  Theorem  the  division  in  this  case  is  not  exact. 

(iv.)  The  binomial  sum  x""  +  y'"  is  or  is  not  divisible  without  re- 
mainder by  the  sum  x  +  y  according  as  m  is  odd  or  even. 

For,  replacing  x  by  (—  y),  we  have  (—  3/)*"  +  y^. 

Examining  this  "remainder"  for  both  odd  and  even  values  oim, 
we  find  that :  ^ 

If  m  be  odd,  (—  yY  +  f  becomes  —  y'^  +  f  =  0. 

Hence  by  the  Remainder  Theorem  the  division  in  this  case  is 
exact. 

If  m  be  even,  (—  yY  -f  y'"  becomes  y"^  -\-f  =  2f  ^0. 

Hence  by  the  Remainder  Theorem  the  division  in  this  case  is  not 
exact. 

66.   By  (i.)  a:"*  —  y"*  is  always  divisible  hy  x  —  y  whether  m  be 


DIVISION  135 

odd  or  even,  and  by  (iii.)  it  is  also  divisible  by  a;  +  ^  when  m  is  even. 
Hence  it  follows  that  when  m  is  even,  £c"*  —  ?/"*  is  divisible  by  either 
X  —  y  ore  x-V  y. 

57.  Since,  when  m  is  even,  the  binomial  difference  m^  —  iT  is 
divisible  by  both  the  ^\xm  x  +  y  and  the  difference  x  —  y/\i  follows 
that  it  is  divisible  by  the  product  {x  +  y)(x  —  y)  =  x"^  —  y\ 

58.  The  general  principles  established  above  may  be  stated  as 
follows  : 

I.  The  sum  of  the  same  odd  poivers  of  two  numheis  is  exactly 
divisible  by  the  sum  of  the  numhers, 

E.  g.  ^  ^  ^  =  a*  -  a%  +  a%'^  -  ab^  +  6*. 

II.  The  difference  of  the  same  odd  powers  of  two  numbers  is 
eixactly  divisible  by  the  difference  of  the  numbers, 

E.  «.  "^  ~  f  =  a«  +  a^b  +  a*b'^  +  a%^  +  a%^  +  ab^  +  b\ 

a  —  0 

III.  The  difference  of  the  same  even  powers  of  two  numbers  is 
exactly  divisible  by  either  the  sum  or  the  difference  of  the  numbers. 

a*  —  b* 
E.  g.  •  (1)         y  =  a»  -  a^  +  ab-^  -  b\ 

(2)         ^^f  =  a^  +  a%  +  ab^  +  6». 

IV.  The  sum,  of  the  same  even  powers  of  two  numbers  is  not 
exactly  divisible  by  either  the  sum  or  the  difference  of  the  two 
numbers. 

The  division  in  each  case  is  not  exact. 


a2 

-\-b'^\ 

a 

+  b 
+  b^ 

a 

-b 

liaw  of  Polynomial  Quotients. 

59.  It  may  be  shown  by  actual  division  that  the  polynomial 
quotients  obtained  by  dividing  a"  ±  6"  by  a  ±  6  have  the  follow- 
ing forms: 


136  FIRST  COURSE  IN  ALGEBRA 

If  w  is  an  odd  positive  integer,  we  have 

The  signs  of  the  terms  of  the  quotient  are  alternately  +  and  — . 
If  71  is  an  odd  positive  integer,  we  have 

^    "^  a  —  o 

The  signs  of  the  terms  of  the  quotient  are  all  positive. 
If  n  is  an  even  positive  integer,  we  have 

(iii.)  ^"7^"  =  a«-i  T  a^-^fr  +  a"-^^^  i^  an-^fts  + +  f,i,«-i  =f  6«-i. 

In  the  expression  above,  the  double  sign  ±,  read  "plus  or  minus," 
is  used  with  the  double  sign  =F,  read  "minus  or  plus,"  to  indicate 
that  according  as  the  upper  or  lower  sign  is  used  in  the  divisor,  the 
corresponding  upper  or  lower  sign  must  be  used  in  the  quotient. 

Hence,  using  the  upper  signs,  the  signs  of  the  quotient  are  alter- 
nately positive  and  negative  when  the  divisor  is  a  sum.  Using  the 
lower  signs,  the  signs  of  the  quotient  are  all  positive  when  the  divisor 
is  a  difiference. 

Mental  Exercise  VIII.    7 

Obtain  each  of  the  following  quotients  mentally,  stating  in  each 
case  the  general  principle  applied  :     (See  §  58.) 

6.  ''•-*' 

7. 
8. 
9. 


5.  J-.  10.  .  xu.  r- 

a  +  0  03+1  a  +  b 


a^+y 

X 

+  y* 

a' 

+  b' 

a 

+  b' 

c« 

-^ 

c 

-d' 

m' 

'-n' 

m 

—  n 

«« 

-b' 

a 

-b' 

a^'  +  b'' 

a 

+  b 

^' 

>_yo 

X 

-y 

a' 

+  8 

a 

+  2* 

x' 

4-  1 

11. 

y^^-^l 

y  -  3 

12. 

\  +  ^ 

l  +  x' 

13. 

l-x' 

1  —X  ' 

14. 

«^2  _  ^,12 

a    —  b 

1 K 

a'-b' 

GRAPHS  137 


CHAPTER  IX 

GRAPHICAL  REPRESENTATION  OF  THE   VARIATION  OF 
FUNCTIONS   OF  A  SINGLE  VARIABLE 

1.  Any  expression  which  depends  for  its  vahie  upon  the  value 
assigned  to  some  specified  variable  contained  in  it  is  called  a 
function  of  this  specified  variable. 

E.  g.  The  expression  x  +  1  does  not  represent  any  particular  number 
until  some  definite  value  is  assigned  to  x.  Hence  we  say  tbut  a:  +  1  i.s  a 
function  of  x. 

Similarly  the  expression  a;^  -f  2  a:  +  3  is  a  function  of  x.  If  x  be  2,  the 
expression  stands  for  the  number  11  ;  if  a;  be  5,  the  expression  represents 
38. 

2.  An  expression  containing  several  variables  may  be  regarded 
as  being  a  function  of  any  one  of  them,  or  of  a  combination  of  two 
or  more  taken  together. 

E.  g.  a;2  4-  a:y  4-  y'^  may  be  regarded  as  being  a  function  of  either  x  ov  y 
separately,  or  of  x  and  y  together. 

3.  From  our  experience  with  algebraic  expressions  we  have  found 
that  definite  numbers  were  commonly  obtained  when  particular 
values  were  assigned  to  the  letters  appearing  in  them.  * 

E.  g.    For  a;  =  2,  tlie  following  functions  of  one  letter  each  represent  7  : 

6a: +  23 


x2  +  3,  a:  +  5,3a:+l,4a;-l  and 


5 


4.  We  found  also  that  a  given  expression  commonly  assumed 
different  values  when  different  values  were  substituted  for  some 
particular  letter  or  letters  appearing  in  it. 

The  expression  x^  —  10  x  -\-  21,  regarded  as  a  function  oix,  will 
represent  different  values  as  x  is  given  successively  the  values  0,  1, 
2,  3,  4,  5, ,  10. 


138 


FIRST  COURSE  IN  ALGEBRA 


We  will  tabulate  the  resulting  values  of  the  function,  writing  the 
results  as  in  the  accompanying  table. 

"When  X  has  the  values  0,  1,  2,  3, ,  the  resulting  values 

of  the  expression  are  21,  12,  5,  0, respectively. 

5.  Writing  the  expression  x^  —  10  a;  +  21  in 
the  form  (x  —3)(x—  7),  it  appears  that,  for 
values  of  x  equal  to  3  or  7,  the  expression  be- 
comes 0.  It  cannot  become  0  for  any  other 
values,  since  neither  of  the  factors  x  —  3  nor 
X—  7  can  become  0  for  other  values. 

For  any  value  of  x  greater  than  7,  the  expres- 
sion will  represent  a  positive  number,  since  each 
factor  jc  ■—  3  and  x—  1  will,  in  this  case,  be  a 
positive  number. 

6.  Also,  for  all  negative  values  of  x,  the  ex- 
pression will  represent  a  positive  number,  since 
for  negative  values  of  x,  x—  3  and  x  —  7  are 
both  negative,  and  their  product  is  accordingly 
a  positive  number. 

We  may  thus,  by  tabulating  results,  obtain  an 
idea  of  the  variation  in  the  value  of  the  expression  under  examina- 
tion as  the  letters  are  given  different  values. 

7.  We  will  now  explain  a  graphic  method  for  representing  the 
variation,  or  change  in  numerical  value,  of  a  given  function  as  the 
letters  appearing  in  it  are  given  different  values. 
i.  We  shall  obtain  a  diagram  or  picture  which  will  represent  to  us 
the  variation  above  and  below  zero  in  the  numerical  value  of  a 
given  function  in  much  the  same  way  as  a  profile  map  of  some 
section  of  a  country  gives  us  at  a  glance  a  different  and  far  better 
idea  of  the  relative  elevations  of  places  than  can  be  obtained  from 
a  table  of  estimated  distances  above  or  below  the  sea  level. 


X 

Function 
a;>-10x-|-21 

0 

21 

1 

12 

2 

5 

3 

0 

4 

-3 

5 

-4 

6 

-3 

7 

0 

8 

5 

9 

12 

10 

21 

Specification  of  Points  in  a  Plane 

8.  We  will  draw  two  perpendicular  straight  lines,  JC'OJT  and 
Y'OY,  as  axes  of  reference  separating  a  plane  into  four  parts 
or  regions.     By  common  consent  among  mathematicia,ns  these  parts 


II 

1 

X' 

III 

r     '" 

GRAPHS  139 

into  which  the  plane  is  separated  are  called  quadrants.  The  parts 
containing  the  Roman  numerals  I,  II,  III,  IV  in  the  accompanying 
diagram  are  called  the  first,  second,  third,  and  fourth  quadrants 
respectively. 

9.  Letting  P  represent  any  point  in  the 
plane,  we  will  suppose  that  lines  such  as  MP 
and  NP  are  drawn  parallel  respectively  to  the 
lines  of  reference  or  axes  Y'O  Y  and  X'OX. 

10.  Starting  at  the  point  of  intersection  0  of 
the  axes  of  reference,  we  may  imagine  taking 
a  first  step  OM  equal  to  NP  along  the  axis  -p      , 
X'OX  from  0  to  M;  then  turning  about  at  M 

and  moving  away  from  the  axis  in  a  direction  MP  parallel  to  the 
other  axis  of  reference  Y'O  Y,  we  shall  arrive  at  the  point  P  by 
taking  a  second  step  from  M  to  P. 

By  taking  steps  of  different  lengths,  we  shall  arrive  at  different 
points  in  the  plane. 

11.  We  shall  call  the  lines  of  reference  X'OX  and  TOY  the 
a?-axis  and  the  |/-axis  respectively,  and  shall  call  their  inter- 
section 0,  from  which  the  first  step  of  any  pair  is  always  taken,  the 
origin. 

12.  We  shall  speak  of  steps  as  being  a;-steps  or  ^/-steps,  ac- 
cording as  they  are  taken  parallel  to  the  a;-axis  or  the  ?/-axis. 

13.  If  we  agree  to  call  steps  taken  along  the  a^-axis  toward  the 
right,  as  indicated  by  the  arrow,  positive  steps,  then  those  toward 
the  left  will  be  negative.  If  y-steps  be  positive  when  taken  upward, 
as  indicated  by  the  arrow,  then  those  downward  will  be  negative. 

14.  Starting  from  the  origin  0,  and  moving  toward  the  right  or 
left,  then  either  up  or  down,  we  may,  by  taking  the  proper  positive 
or  negative  a;- steps  and  ?/-steps,  reach  points  lying  in  any  of  the 
four  quadrants. 

15.  If,  starting  from  0,  we  first  take  a  positive  ai-step  and  then 
take  a  ^/-step,  we  shall  pass  into  either  the  first  or  the  fourth 
quadrants,  according  as  the  ?/-step  is  positive  or  negative. 

If  our  first  ic-step  be  negative,  then  the  y-step  which  follows  it 
will  take  us  into  either  the  second  or  the  third  quadrants. 

Hence  the  signs  of  the  steps  which  may  be  taken  to  reach  a  point 
determine  the  quadrant  in  which  it  lies. 


140 


FIRST  COURSE   IN   ALGEBRA 


II 


III 


(3.2) 


X 
(3.-2) 


16.  With  reference  to  any  particular  point,  i?,  the  corresponding 
ic-step  is  called  the  abscissa,  from  the  Latin  meaning  "  to  cut  off," 
and  the  y-step  is  called  the  ordiuate,  from  the  Latin  meaning  "  to 
set  in  order  or  arrange." 

17.  Since  the  abscissa  and  ordinate  taken  together  enable  us 
to  locate  any  particular  point,  they  are  together  called  the  co- 
ordinates of  any  particular  point,  and  the  axes  parallel  to  which 
they  are  drawn  are  called  the  coordinate  axes. 

18.  Whenever,  in  the  system  of  graphic  representation  which 
we  are  presenting,  the  position  of  a  point  is  described  by  means  of 
its  coordinates,  the  a-step  or  abscissa  is  understood  to  be  the  one 
named  first,  unless  the  contrary  is  stated. 

19.  By  the  notation  (3,  2)  we  shall  under- 
stand that  the  abscissa  of  the  point  repre- 
sented is  3  and  its  ordinate  2.  After  assuming 
some  convenient  unit  of  length,  we  may  locate 
the  point  by  first  measuring  off  a  distance  of 
three  units  irom  the  origin  0  along  the  jc-axis 
toward  the  right.  Then,  turning  about  in  a 
direction  parallel  to  the  ^/-axis,  we  shall  find 

the  point,  P,  situated  at  a  distance  of  two  units  measured  in  a 

positive  direction,  that  is,  upward.     (See  Fig.  2.) 

20.   The  point  (3,  —  2)  is  situated  in  the  fourth  quadrant  at  a 

distance  of  three  units  to  the  right  of  the  2/-axis,  and  two  units 

below  the  axis  of  X,  as  in  Fig.  2. 

21.  Such  points  as  those  represented 
by  (-  4,  -I-  3),  (-  4,  -  2),  (0,  2),  ((J,  -  1), 
(2,  1),  (2,  -  3;,    (3,  3),  (4,  0),  etc.,  will 

(2,1)  i  be  readily  located  by  stepping  off  the  in- 

1  '■  ^^^  dicated  distances,  first  toward  right  or 
left  fi-om  the  origin,  then  up  or  down,  ac- 
cording as  the  signs  of  the  given  coordi- 
nates are  +  or  — .     (See  Fig.  3.) 

22.  The  scale  units  in  terms  of  which 
Fig.  3.                  the  a;-steps  and    ?/-steps  are    expressed 

are  understood  to  be  the  same  unless  the  contrary  is  stated. 

It  may  happen  for  special  purposes  that  it  is  convenient  to  choose 


IV 

Fig.  2. 


II 

,(-4.3) 


l(-4.-2) 


III 


Y     I 


(0,2) 


t(3.3) 


(0-1)  I 


1(2,-3) 


IV 


aRAPHS 


141 


a   scale  unit  for  the  y-steps  different  from  that  chosen  for  the 
jc- steps. 

23.    The  operation  of  marking  any  point  on  a  diagram  when  the 
coordinates  of  the  point  are  given,  is  called  plotting  the  point. 

Exercise  IX.     1 
Plot  the  points  whose  coordinates  are 


1.   (2,  4). 

5.  (1,  -  5). 

9.   (4,  -  4).        13.  (3,  0). 

2.   (8,  6). 

6.  (2,  -  4). 

10.   (-  2,  4).        U.   (0,  2). 

3.   (1,  7). 

7.   (3,  -  1). 

11.  (0,  -  4).        15.   (0,  0). 

4.   (5,  8). 

8.  (-  3,  11). 

12.   (-  1,  -  1).   16.   (-  6,  -  6). 

.(1.12) 


1 

I 
1    I 


I 

I  t(2.5) 

I  I 

I  t 

i  I 

I  I 

I  I 

I  I 


(8.5), 


(9.12) 


24.  If  we  regard  the  sets  of  values  in  the  table  calculated  from 
the  function  x^—U)x  +  2l  (see  §§  4,  5)  as  representing  the 
ic-coordinates  and    ^/-coordinates   of 

different  points,  we  may  obtain  as 
many  points  in  a  plane  as  we  please, 
such  as  (1,  12),  (2,  5),  (3,  0),  etc. 
We  shall  assume  the  numbers  in 
the  column  under  x  as  abscissas, 
and  those  in  the  column  under 
.2^^—10^+21  as  ordinates,  as  in 
Fig.  4. 

25.  By  assigning  fractional  values 
to  X  we  may  obtain  corresponding 
values  of  the  function. 


0 


(3.0)1 


1  (7.0) 


E.  g.    If  a;  be  given  values  between  0 
and  1,  say  .1,  .2,  .3,  .4,  etc.,  we  may  cal-  (4,-3)     .     (6,-3) 

dilate  as  corresponding  valnes  of  y,  20.01,  (5,-4) 

19.04,  18.09,  17.16,  etc.  ,Fig.  4. 

The    points    corresponding    to    these 
values  taken  as  coordinates  will  be  found  to  lie  between  tlie  points  (0,21) 
and  (1,12). 

26.  Continuity*  If  we  imagine  that  the  value  of  x  changes  con- 
tinuously from  0  to  1,  then  from  1  to  2,  and  on  to  3,  4,  5,  etc.,  pass- 
ing through  all  intermediate  values  without  at  any  stage  making  a 
sudden  "jump"  from   one  value  to  another,  then  the  point  P, 


142 


FIRST  COURSE  IN  ALGEBRA 


determined  by  x  and  y  (Fig.  2)  will  trace  out  a  continuous  line,  — 
that  is,  one  having  no  breaks  or  discontinuities  in  it. 

27.  A  continuous  line  passing  through  all  of  the  points  which 
may  be  plotted  for  a  given  function  is  called  the  graph  of  the 
function. 

28.  The  graph  of  a  given  function  is,  from  its  construction,  a 
"  point  picture  "  of  an  algebraic  expression. 

A  portion  of  the  graph  of  the  function  ar*— 10a;  +  21  will  be 
found  by  drawing  a  continuous  line  through  the  points  plotted  in 

Fig.  4.  The  "  accuracy  "  of  the  graph, 
|{ai2)  |.jjg^^  -g^  ^jjg  <c  likeness  of  the  picture," 
will  depend  largely  upon  the  lengths 
of  the  scale  units  adopted  in  its  con- 
struction.    (See  Fig.  5.) 

29.  The  equality  y  =  x^—\^x-\- 
21,  or  in  general,  y  =f(x),  is  called 
the  equation  of  the  graph  or 
curve. 

To  every  pair  of  values  satisfying 
the  equality  y  =  f{x)  there  corre- 
sponds a  point  on  the  graph;  and 
conversely,  the  coordinates  of  all 
points  on  the  graph,  and  no  other 
points,  satisfy  the  equation. 

30.  Whenever  we  construct  a  graph 
by  drawing  a  continuous  line  through  different  points  which  are 
separately  plotted,  we  assume  in  the  operation  that  the  graph  of 
the  function  is  continuous  between  any  two  points  through  which 
the  continuous  line  passes. 

We  need  not,  at  present,  be  at  all  concerned  with  the  subject  of 
"  breaks  "  or  discontinuities  in  the  graph,  since  it  may  be  proved 
that  the  graph  of  every  rational  integral  function  of  one  variable  has 
no  discontinuities.  It  may  also  be  shown,  for  finite  values  of  the 
variable,  that  a  rational  fractional  function  has  no  discontinuities 
except  for  such  values  of  the  variable  as  make  the  denominator  zero. 

31.  As  an  illustration  of  a  break  or  discontinuity  in  a  graph  or 
curve  see  the  accompanying  figure  where  we  suppose  that  the  point 


GRAPHS 


143 


which  is  tracing  the  curve,  on  arriving  at  P,  jumps  immediately  to 
Q  through  a  finite  distance  FQ,  and  then  moves  on  along  the  branch 
of  the  curve  QQ\  y] 

Typical  Graphs 

32.    Consider  a  function  of  the  form  «ic^,  in    

which  a  is  positive. 

By  assigning  some  numerical  value  to  a  through- 
out the  discussion,  we  may,  by  giving  different    ""^ 
values  successively  to  x,  calculate  the  corresponding 
values   of  the  function  ax^.     If  we  represent  the 
function  by  ?/,  we  may  write  y  =  ax^. 
r 


Fig.  0. 


Fig.  7. 


value  of  the 

If,  for  the  present  purpose, 

we  let  a  =  2,  we  may  obtain  the  graph  of  ?/  =  2  aj^ 

by  plotting  the  points  whose  abscissas  and  ordinates 

are  given  by  x  and  2  x^  respectively. 

33.  Since  the  number  represented  by  a  is  positive 
and  x^  is  an  even  power,  ax"^  will  be  positive  for  all 
values  of  x.  It  follows  that  there  can  be  no  negative 
ordinates,  and  the  graph  can  enter  neither  the  third 
nor  the  fourth  quadrants.  A  portion  of  the  graph  of 
the  function  2  x^  which  has  the  form  ax^  is  exhibited 
in  Fig.  7  in  full  lines,  and  we  have  shown  in  dotted 
lines  on  the  same  diagram  a  portion  of  the  graph  of 
another  function,  4ic^,  whose  algebraic  form  is  the 
same  as  that  of  the  function  2  x^. 

It  will  be  seen  that  the  graphs  have  the  same 
general  "  shape "  except  that  one  comes  to  a 
"sharper"  point  than  the  other. 

34.  In  case  the  number  represented  by  a  is  nega- 
tive, that  ,is,  if  in  particular  a  represents  —  2,  the 
ordinates  corresponding  to  the  function  —2x^  will 
be  negative  for  both  positive  or  negative  values  of  x. 
Hence,  no  part  of  the  graph  of  —  2  a;^  can  enter  either 
the  first  or  the  second  quadrants.  The  graphs  of 
the  two  functions  2  x^  and  —2x^  having  the  form 
ax\  but  in  one  of  which  the  number  represented  by 
a  is  positive  and  in  the  other  negative,  will  have 


144         FIRST  COURSE  IN  ALGEBRA 

the  same  shape  and  size,  but  will  be  situated  symmetrically  with 
respect  to  the  axis  of  X  as  in  Fig.  8. 

35.  If  we  imagine  the  ic-axis  to  be  a  plane  mirror  perpendicular 
to  the  ^-axis,  either  graph  may  be  regarded  as  being  the  reflection 
of  the  other  in  the  ic-axis  as  a  mirror. 

By  interchanging  the  letters  x  and  y,  we  may,  from  the  expression 
of  equality  y  =  aa^,  obtain  x  =  ai/^.  Hence,  when  plotting  this  last 
function,  x  =  at/^  we  may  simply  interchange  the  values  represented 
by  X  andy,  and  make  the  same  measurements  as  before  for  the 
function  y  =  ax\ 

36.  Interchanging  these  values  amounts  to  revolving  the  entire 
system  toward  the  right  about  the  origin  as  a  center,  keeping  the 

axes  and  graph  in  the  same  relative  po- 

I  sitions,   through  an  angle  such  that  the 

j  y-axis  shall  swing  around  into  a  position 

'  originally  occupied  by  the  ic-axis.     Hence, 

/  if  in  Fig.  9  the  dotted  line  represents   a 

portion  of  the  graph  of  2  x^  in  its  original 

^"^  position,  and  we  imagine  the  system  to  be 

» JC    revolved  about  the  origin  0  until  0  Y  comes 

^ into   the  position   originally   occupied   by 

^'^-  ^-  OX,  the  graph  may  be  supposed  to  turn 

about  and  appear  in  a  position  indicated  by  the  full  line.     It  may 
then  be  taken  as  the  graph  of  the  function  2  if. 

37.  Consider  now  the  function  ax^  +  h.  To  ob- 
tain the  graph  of  this  function  we  have  simply  to  add 
b  to  each  of  the  ordinates  calculated  for  the  graph  of 
ax^.  Hence,  the  ^-coordinates  will  be  the  same  for 
both  of  the  functions  ax^  and  ax^  +  h,  but  the  y-ordi- 
nates  of  the  function  am^  +  b  will  be  greater  by  b 
than  those  of  the  function  ax^.  It  follows  that  the 
two  graphs  have  the  same  size  and  shape,  but  because 
of  the  greater  length  of  the  ?/-ordinates  the  graph  of 
the  function  ax^  +  ^  is  situated  "  higher  up  "  in  the 
plane  than  the  graph  of  the  function  ani?.     That  is, 

it  will  be  farther  away  from  the  origin  in  the  ?/- direction,  as  in  Fig. 
10.     In  this  figure  we  have  given  a  and  b  the  values  2  and  3  respect- 


«  \ 

y  /  1 

\\ 

/ 1 

\  \ 

/  1 

\  \ 

/  1 

\  \ 

/  1 

\  \ 

/  1 

\  \ 

/  1 

\ 

\ 

1 

\ 

1 

\ 

/ 

\ 

/ 

\ 

/ 

,  \. 

y  , 

0   X 

GRAPHS 


145 


ively,  and  have  shown  a  portion  of  the  graph  of  the  function  2x^+3 
in  full  lines  and  a  portion  of  the  graph  of  the  function  2  x^  in 
dotted  lines. 

38.  In  Fig.  1 1  we  have  shown  portions  of  the 
graphs  of  (i.)  2  x\  (ii.)  2  cc^  +  3,  and  (iii.) 
x^  +  2x  —  ^. 

It  may  be  noted  that,  so  far  as  the  portions  of 
the  graphs  shown  are  concerned,  they  have  in 
all  cases  the  same  general  shape. 

39.  We  will  now  show  representative  graphs 
of  certain  common  mathematical  curves,  leaving 
the  student  to  plot  the  functions  in  the  next 
exercise,  comparing  the  forms  of  the  graphs 
obtained  with  those  shown  in  Figures  12  to  16. 

We  cannot  always  safely  assign  a  name  to  a 
curve  simply  because  a  certain  portion  of  it  re- 
sembles very  closely  the  shape  of  a  curve  whose  name  is  known,  but 
we  may  at  least  assert  that  the  portions  of  the  curves  obtained 
appear  to  have  the  same  general  form. 


Fig.  11. 


Fig.  13.  Hyperbola 


Fig.  U..  Ellipse 


Fig.  15.  Cubic  Curve 


Fig.  16.  Cubic  Curve 


40.    The  student,  after  having  constructed  a  few  of  the  simple 
typical  graphs,  will  understand  the  term  "  algebraic  form  "  in  a  new 

10 


146  FIRST  COURSE  IN  ALGEBRA 

light.  It  will  be  seen  that  there  is  a  striking  similarity  in  shape 
among  the  "  point  pictures  "  or  graphs  of  algebraic  expressions  which 
have  the  same  algebraic  form. 

41.  Although  the  number  of  different  graphs  which  may  be  con- 
structed by  plotting  functions  is  as  endless  as  the  number  of  ex- 
pressions themselves  which  may  be  written,  comparatively  few  of 
these  forms  will  be  encountered  in  elementary  work,  and  these  can 
readily  be  separated  into  a  certain  small  number  of  typical  forms 
which  may  be  easily  recognized. 

We  present  a  few  of  the  more  common  forms  in  order  that  the 
student  may  become  somewhat  acquainted  with  certain  of  the  graphs 
which  are  of  practical  and  historic  interest. 

Exercise  IX.     2 
Obtain  portions  of  the  graphs  of  the  following  equations  : 


1.  y  =  x. 

23. 

1 

2.  y  =  —  x. 

^      x-h  I 

3.  y  =  2ic. 

4.  y  =  —  4:X. 

24. 

1 
''  =  x-l' 

5.  x  =  2. 

6.  y  =  S. 

25. 

1 
^      x+2 

7.  2/  =  0. 

26. 

x^ 

8.  x  =  0. 

^-x+l' 

d.  y  =  x-\-l. 

27. 

xy  =  24. 

10.  y  =  -x+l. 

28. 

xy  =  ~l. 

11.  y  =  Sx  +  4.. 

29. 

y  =  x^-~9x^+  nx+  21. 

12.  y  =  —  2x+  5. 

30. 

y  =  X^  —  4:X. 

13.  y  =  4.x\ 

31. 

y  =  x^-Sx^+2Sx  +  2. 

14.  y  =  x\ 

32. 

y  =  x^  +  x^  +  x+  1. 

15.  y  =  x^+2x-\-  1. 

33. 

y::=x^  -\-Sx^  +  2x+  1. 

16.  y  =  x'+lx-\-12. 

34. 

y  =  x^-Sx^+  2a;- 1. 

17.  y  =  x^-lx+d. 

35. 

y  =  —x^+3x''  —  2x+h 

18.  y  =  x^-6x  +  9. 

36. 

y  =  a;8  +  1. 

19.  y  =  -4:x'+20x~23. 

37. 

7/  =  £c«  +  3a;'+  6a;+  11. 

20.  y  =  —  2x^-\-x. 

38. 

y  =  x^—Sx''  ■]-  Qx—  11. 

21.  y  =  1- 

39. 
40. 

y  =  x^  +  4.x''+2x  +  5. 
y  z=  x^  —  4:  a^  -\-  2  £c  —  5. 

^^-y  =  i' 

41. 

y  =  x^-10x^-{-  12. 

GRAPHS 


147 


Inverse  Use  of  Graphs 

42.  When  constructing  graphs  we  have  assumed  that,  starting 
from  the  origin  0,  every  point  in  the  plane  may  be  reached  by 
taking  definite  ic-steps  and  ?/-steps.  It  may  be  seen  that,  when 
the  location  of  a  particular  point  is  given  relatively  to  the  axes  of 
reference,  we  may  by  reversing  the  process  use  a  scale  and  obtain 
by  actual  measurement  approximate  values  for  the  cc-coordinate  and 
the  ^/-coordinate  of  the  point. 

The  accuracy  of  the  numerical  results  thus  obtained  will  depend 
upon  the  accuracy  with  which  the  measurements  are  made. 

43.  The  graph  of  a  function  is  obtained  by  assigning  values  to 
the  variable  appearing  in  it,  and  by  locating  points  whose  coordi- 
nates are  the  values  assigned  to  the  variable  and  the  corresponding 
values  calculated  for  the  function. 

If  the  graph  of  a  function  be  given,  we  may  measure  the  coordi- 
nates of  different  points  situated  on  it  and  estimate  approximately 
the  values  which  must  be  given  to  the  variable  in  order  that  the 
function  shall  have  certain  specified  values. 

44.  We  may  find  an  approximate  value  for  VY  by  using  the  graph 
of  cc  =  vy  as   follows  : 

Since  like  powers  of  equal  expressions  are  equal,  from  x  =  Vv/  it 
follows  that  ar^  =  y. 

A  portion  of  the  graph  oix^  =  y  may  be  obtained 
by  using  as  abscissas  different  values  assigned  to 
£c,  and  for  ordinates  the  corresponding  values  cal- 
culated for  the  function  ar^. 

When  to  x  is  given  successively  the  values  0,  1, 
2,  8,  4,  etc.,  the  corresponding  values  of  the  func- 
tion x^  are  0,  1,  4,  9,  16,  etc.,  as  shown  in  the 
accompanjdng  table. 

By  using  these  pairs  of  values,  (0,0), (1,1), (2,4), 
(3,9),  (4,16),  etc.,  as  abscissas  and  ordinates,  points 
upon  the  graph  may  be  located.     (See  Fig.  17.) 

The  accuracy  of  the  figure  obtained  by  drawing  a  continuous 
line  through  the  points  thus  located  will  depend  largely  upon  the 
lengths  of  the  scale  units  employed  when  setting  off  the  abscissas 
and  ordinates. 


Abscissa 

Ordinate 

X 

x^=y 

6 

6 

1 

1 

2 

4 

3 

9 

4 

16 

148 


FIRST  COURSE  IN  ALGEBRA 


To  find  Vt  our  problem  is  simply  to  find  the  abscissa  of  a  point 
on  the  graph  corresponding  to  the  ordinate  whose  length  is  7.  To 
do  this  we  may  start  at  the  origin  and  move  upward  along  the 
y-axis  through  a  distance  equal  to  seven  of  the  scale  units  used  in 
constructing  the  graph.  Then,  tracing  along  the  horizontal  line 
passing  through  this  point  to  the  graph,  we  shall  find  the  point 
whose  ordinate  has  the  required  length,  7. 

The  Vt  which  is  represented  by  the 
length  of  the  abscissa  a\  corresponding  to 
the  point  thus  located  on  the  graph,  may 
be  readily  estimated  by  measuring  the  dis- 
tance passed  over  in  tracing  across  from 
the  axis  of  Y  to  the  graph.  Thus,  fi-om 
Fig.  17,  in  which  each  small  square  repre- 
sents .2,  the  value  of  Vl  is  found  to  be 
2.6 -I-,  nearly. 

45.  Approximate  values  for  the  square 
roots  of  other  numbers,  such  as  2,  3,  5,  6, 
8,  etc.,  may  be  found  in  a  similar  way  by 
tracing  across  the  figure  from  the  axis  of 
Y  to  the  graph  along  horizontal  lines  situ- 
ated at  distances  of  2,  3,  5,  6,  8,  etc.,  units 
respectively,  above  the  ;r-axis. 

We  may  thus  obtain  as  approximations 
for  the  square  roots  of  2,  3,  5,  6,  8,  etc., 
the  numbers  1.4 -I-,  1.7 -f,  2.2+,  2.5+, 
2.8  +,  etc. 


IDENTICAL  EQUATIONS  149 


CHAPTER  X 

GENERAL   PRINCIPLES   GOVERNING  TRANSFORMATIONS  OF 
ALGEBRAIC   EQUATIONS 

Identical  Equations 

1.  An  equality  or  equation  is  the  assertion  in  s)mibols  that 
two  different  expressions  represent  the  same  number. 

2.  Two  expressions  are  said  to  be  identical  if  they  are  exactly 
alike,  or  if  either  can  be  reduced  to  the  form  of  the  other  by  apply- 
ing the  laws  of  reckoning  and  the  definitions  of  the  symbols  and 
functions  considered. 

E.  g.  The  expressions  a  +  b  and  a  +  b  are  identical  since  tliey  are  ex- 
actly alike. 

The  expressions  2  a  +  3  a  and  5  a  are  identical  since  2  a  -{-3  a  may  by 
addition  be  reduced  to  5  a. 

Since  the  expression  (a  +  b)c  may  l)y  multiplication  be  transformed  into 
the  expression  ac  +  be,  it  follows  that  (a  +  b)c  is  identical  with  ac  -\-  be. 

3.  Two  numerical  expressions  which  represent  the  same  number 
are  said  to  be  identical. 

E.  g.  Since  7  +  5  and  4x3  each  represent  12,  it  follows  that  7  +  5  and 
4x3  are  identical. 

4.  The  triple  sign  of  equality,  = ,  commonly  called  the  identity 
Migii,  —  which  is  read,  "is  identical  with,"  "may  be  transformed 
into"  or  "becomes,"  —  is  written  between  two  expressions  to  de- 
note that  they  are  identical. 

5.  In  an  identity  the  expression  at  the  left  of  the  identity  sign 
is  called  the  firsst  member,  and  the  expression  at  the  right  of  the 
sign  the  second  member. 

E.  g.  In  the  identity  2  a  +  3  «  =  5  a,  the  expression  2  a  +  3  a  is  the  first 
member  and  5  a  is  the  second  member. 


150  FIRST  COURSE  IN  ALGEBRA 

6.  Since  either  member  of  an  algebraic  identity  may  be  reduced 
to  the  form  of  the  other,  it  follows  directly  that,  if  both  members 
are  finite  expressions^  they  must  represent  equal  numerical  values  for 
all  values  of  the  letters  which  appear  in  them. 

E.  g.  Since  {a  +  by  can  be  expressed  in  the  form  a^  +  2  ah  +  h%  it 
follows  that  («  4-  6)-^  =  a^  +  2  a6  +  h^  lor  all  values  which  may  be  assigned 
to  a  and  b. 

7.  In  dealing  with  identical  equations  we  are  governed  by  the 
following 

General  Principle:  Either  member  of  an  algebraic  identity 
may  be  reduced  directly  to  the  form  of  the  other ,  or  both  members  may 
be  reduced  to  a  common  third  formy  by  applying  the  principles  and 
definitions  of  algebra. 

8.  In  applying  this  principle  it  will  in  certain  cases  be  found 
convenient  to  transform  one  member  of  an  algebraic  identity  directly 
to  the  form  of  the  other. 

E.  g.  In  the  identity  (a:  +  3)  (a:  +  2)  =  a:^  +  5  ^  +  6,  the  first  member 
may  be  reduced  directly  to  the  form  of  the  second  by  performing  the  indi- 
cated multiplication,  and  in  a  later  chapter  a  method  will  be  shown  for 
reversing  the  process  and  reducing  the  second  member  to  the  form  of  the 
first. 

In  some  cases  the  two  members  of  an  algebraic  identity  may  be  of 
such  forms  that  one  does  not  readily  reduce  to  the  form  of  the  other. 
Hence,  in  such  cases,  it  is  convenient  to  transform  both  members  to 
a  common  third  form. 

a4  _  ^4 

E.  g.    In  the  identity  (a  +  &)2  —  2  a6  =  -^ j^ »  ^^'^  shall  by  performing 

the  indicated  operations  in  the  members  separately,  obtain  in  each  case  the 
expression  a^  +  h^. 

Accordingly,  by  reducing  the  members  of  the  given  identity  to  the  com- 
mon third  form,  a^  +  b\  we  shall  derive  the  identity  a"^  -\- b^  =  a^  -[-  b^. 

9.  An  equality  may  be  proved  to  be  an  identity,  either  by 
showing  that  one  member  may  be  reduced  directly  to  the  form  of  the 
othery  or  by  shewing  that  both  members  may  be  reduced  to  a  common 
third  form  for  all  values  which  may  be  given  to  the  letters  which 
appear  in  them. 


IDENTICAL  EQUATIONS  151 

Mental  Exercise  X.     1 

Show  that  each  of  the  following  equalities  is  an  identity  : 

1.  2«  +  Sa  =  5a.  15.  23//'  +  5b^=lb  X  4.b. 

2.  AO  +  (jb  =  106.  16.  3ic^  —  dc^=  lie  X  2c. 

3.  8  c  —  2  c  =  G  c.  11.  (x  +  2)(x  —  2)  =  x"  -  4.. 

4.  llc/=6(^+ 5c?.  18.  (i/+ S)(i/-'d)=f~d. 

5.  lSe  =  20e  —  le.  19.  (;2  +  8) (;s  -  8)  =  ;j' -  64. 

6.  6  7W  +  9  7W  =  4  7«  +  11  w.  20,  (1  +  m)(l  —  m)  =  I  —  m\ 

7.  7  ??  +  lOw  =  20w  —  3?^.  21.  (a  +  3)'  =  a=^  +  6«  +  9. 

8.  12a;—  4cc=  17a;-  9ic.  22.  (6  +  bf  =  36  H-  126  +  6^. 

9.  2(3  r*  +  4  6)  =  6  a  +  8  b.  23.  (9  —  c)^  =  81  —  18  c  +  c^ 

10.  3(56+7c)=156+21c.  24.  ^r  (a  +  4)  +  4  =  ^  2_|_  4^^+ 1), 

11.  4(6c-5)  =  24c-20.  25.  6H  16(6+4)=/>(6+ 16)+64. 

12.  13a  +  7a  =  (5  X  4>.  26.  a;(a!-20)+100  =  a;2— 20(a.'-5). 

13.  166  +  36  =  386-^2.  27.  a;(a!-6)  +  3  (2a;-3)  =  a;'''-9. 

14.  Ida^  +  5a^~Sa  X  3a.  28.  «(;«  —  12)  +  12(a;— 3)=a;^- 36. 

29.  x(x  +  18)  -  9  (2a3  +  9)  =  x^  -  81. 

30.  a;(a;  -  22)  +  11  (2  a;  —  11)  =  a;^  -  121. 

31.  x{x  +  24)  -  24  (a;  +  6)  =  ar^  —  144. 

32.  28  (a;  +  7)  -  x(x  +  28)  =  196  -  x\ 

33.  {a+  iy-4.a  =  {a-  1)\ 

34.  (6  +  2)^^-86^(6-2)1 

35.  (a  +  3)'^  -  12  a  =  (a  —  3)'. 

36.  (c  +  5)2  -  20  c  =  (c  -  5)-". 

37.  (d-Ay+  lGd  =  (d  +  4.y. 

38.  (m  -  6)2  +  24  W2  =  (tw  +  6)1 

39.  (7  -  ny  +  28  w  =  (7  +  /?)'. 

40.  (2  a  +  by  —  Sab  =  (2  a  —  6)1 

41.  (36-46-)2  + 48  6c=  (3  6  + 4c)2. 

42.  (a+  1)2 -(a-  l)2  =  4a. 

43.  (6 +  2)2 -(6 -2)2  =  86. 

44.  (c  +  .5)2  -  (c  -  5)2  =  20  c. 

45.  (7-a;)2-(7  +  a;)2  =  -28a;. 

46.  (9-y)2-(9  +  2/)'  =  -36  2/. 

47.  (a  -  5 6)2  _  (a  +  5 6)2  =  -  20a6. 

48.  (a  +  6)2  +  (a  -  6)2  =2^2  +  262. 


152  FIRST  COURSE  IN  ALGEBRA 

49.  (b  +  cy  +  (/>  -  c)-  =  2  />2  +  2  cl 

50.  (m  -  ny  +  (m  +  ny  =  2  (w-  +  n^, 

51.  (1  -  xy  +  (l+xy  =  2  +  2  x\ 

52.  aj(a;  +  y)  +  i/-^  =  ic^  +  i/(ic  +  y). 

53.  «(«  -  6)  +  6^  =^2  -  h{a  -  b). 

54.  x^  +  2/(2/  —  ic)  =  i»(ic  —  2/)  +  i/^ 

55.  a6  +  c  (rt  +  6)  =  b{a  +  f )  +  «c. 

56.  «(6  +  c)  -\-  bc  =  c{a  +  ^)  -h  ab. 

57.  a*?/  +  -(«  —  y)  —xz  +  y{x  —  z). 

58.  a(6  +  r)  —  cC^r  +  b)  =  b{a  —  c). 

59.  x(y  -  z)  +  z(x  +  y)=  y(x  +  z). 

60.  x{z  +  ?/')  +  y{z  +  ?/')  =  z{x  -\-  y)  +  w{x  +  2/). 

61.  x^  +  {x'  +  xy-^-  y^)y  =  xix"  +  a-//  +  y")  +  y\ 

62.  jc(y  -  ~)  +  y{z  -x)  +  z{x  -  y)  =  0. 

Conditional  Equations 

10.  The  number  which  an  expression  such  as  cc  +  3  may  repre- 
sent depends  entirely  upon  the  particular  value  which  may  be 
given  to  x. 

E.  g.  If  X  be         1,  then  x  +  3  represents      4. 

If  X  be        6,  then  a:  +  3  represents      9". 
If  X  be        0,  then  x  +  3  represents      3. 
If  X  be  —  10,  then  x  +  3  represents  —  7. 
etc.  etc. 

It  should  be  understood  that,  unless  some  condition  is  imposed  which 
restricts  x  to  some  particular  value,  the  expression  x  +  3  taken  by  itself 
may  represent  any  number  whatever. 

If  an  expression  such  as  x  +  3  be  assumed  to  represent  some  particular 
number,  such  as  5,  it  may  be  seen  that  a  restriction  is  placed  u|)on  the 
value  of  X  by  this  assumed  condition.  We  shall  lind  that  this  condition  is 
satisfied  providing  x  is  restricted  to  the  value  2. 

11.  Equalities  which  are  true  only  on  condition  that  specified 
letters  or  sets  of  letters  appearing  in  them  be  given  definite  values 
or  sets  of  values,  are  called  conditional  equations. 

12.  In  a  conditional  equation,  the  expression  at  the  left  of  the 
sign  of  equality  is  called  the  first  member  of  the  equation  and 
the  expression  at  the  right  of  the  sign  the  second  member. 


CONDITIONAL  EQUATIONS  153 

13.  The  terms  of  the  algebraic  expressions  which  form  the  mem- 
bers of  a  conditional  equation  are  called  the  terms  of  the  equation. 

14.  In  order  to  distinguish  conditional  equations  (which  are  true 
for  particuktr  values  only  of  the  letters  appearing  in  them)  from 
identities  (which  are  true  for  all  values  of  the  letters),  we  shall  use 
the  double  sign  of  equality,  =,  when  writing  conditional  equations, 
and  the  triple  sign  of  equality,  = ,  when  writing  identical  equations 
in  which  one  or  more  letters  appear. 

When  writing  numerical  identities,  that  is,  identities  in  which 
numbers  alone  appear,  we  shall  use  the  double  sign  of  equality,  =. 

All  definite  arrangements  of  number  symbols  used  in  arithmetic 
to  represent  numbers  (except  such  as  contain  zero  as  a  divisor) 
represent  definite  numerical  values.  Hence,  when  two  such  arrange- 
ments of  symbols  are  written  as  members  of  a  numerical  equality, 
no  condition  affecting  the  value  of  either  can  exist.  That  is,  a 
numerical  equality  is  always  an  identity. 

The  use  of  the  sign  =  for  numerical  identities  in  algebra  conforms 
with  the  use  of  this  sign  in  arithmetic. 

15.  Although  two  different  algebraic  expressions  may  represent 
unequal  numbers  when  numerical  values  are  given  to  the  letters 
appearing  in  them,  it  may  happen  that  they  represent  equal  num- 
bers on  condition  that  some  particular  value  or  set  of  values  is 
assigned  to  certain  specified  letters  appearing  in  them. 

E.  g.  It  may  be  seen  that  the  two  expressions  2  a;  +  5  and  a?  +  8  represent 
unequal  numbers  when  x  is  given  the  values  I,  2,  or  5. 

If  a;  be  1,  then  2  a;  +  5  represents  7,  and  x  +  ^  represents  9. 

If  X  be  2,  then  2  a;  +  5  represents  9,  and  a:  +  8  represents  10. 

If  X  be  5,  then  2x  +  5  represents  15,  and  a;  +  8  represents  13, 

On  condition  that  x  be  given  the  particular  vahie  3,  tlie  expressions 
2  a;  +  5  and  a;  +  8  each  represent  11,  and  accordingly  we  may  construct  the 
conditional  equation  2a:  +  5  =  a;  +  8. 

The  equality  4a;  —  7  =  2a:  +  3  is  a  conditional  ecpiation  in  which  the 
members  represent  the  number  13  on  condition  that  x  be  given  the  particular 
value  5. 

16.  A  conditional  equation  is  said  to  be  integral,  fractional, 
rational,  or  irrational,  with  respect  to  certain  specified  letters 
appearing  in  it,  according  as  the  algebraic  terms  appearing  in  its 


154  FIRST  COURSE   IN   ALGEBRA 

members  are  integral,  fractional,  rational,  or  irrational  with  respect 
to  these  letters. 

E.  g.  The  conditional  eqiuation  a:*  +  7  a:^  =  4  is  integral  and  rational 
with  respect  to  x,  since  x  docs  not  appear  in  the  denominator  of  a  fraction 
in  any  term,  nor  under  a  radical  sign. 

The  conditional  equation =  x  —  3  is  fractional  witli  reference  to  x, 

^  a;  +  1  ' 

since  x  appears  in  the  denominator  of  the  fraction  in  the  first  member.     It 

is  also  rational  with  reference  to  x,  since  x  does  not  appear  under  a  radical 

sign.  

The  conditional  equation  \/x  +  2  +  a:  =  10  is  irrational  with  reference 

to  a:,  since  x  appears  under  the  radical  sign  in  the  first  term. 

17.  The  degree  of  a  conditional  equation  which  is  integral  and 
rational  with  respect  to  one  or  more  specified  letters  is  equal  to  the 
degi*ee  of  the  term  of  highest  degree  with  reference  to  these  letters. 

E.  g.  Tlie  conditional  equation  ar*  +  2  a:^  —  5  a;  +  1  =  0  is  of  the  third 
degi-ee  with  reference  to  .r. 

The  conditional  equation  x^y  +  2  a;  —  i/^  =  9  is  of  the  third  degree  with 
reference  to  x  and  y  together. 

It  should  be  observed  that  the  definition  given  for  the  degree  of 
a  conditional  equation  requires  that  the  equation  be  neither  frac- 
tional nor  irrational  with  reference  to  the  letters  in  terms  of  which 
its  degree  is  reckoned. 

Equations  which  are  either  fractional  or  irrational  with  reference 
to  certain  letters  appearing  in  them  are  not  spoken  of  as  having 
degree. 

E.  g.   The  conditional  equations =  x  —  3  and   ^x  +  2  +  a;  =:  10 

cannot  be  considered  as  having  degree. 

18.  Since,  when  a  conditional  equation  is  constructed,  we  may 
not  know  the  value  or  values  which  must  be  assigned  to  a  specified 
letter  or  set  of  letters  appearing  in  it  in  order  that  the  expressed 
equality  may  be  true,  it  is  consistent  to  speak  of  these  letters  as 
unknowns. 

E.g.  In  the  conditional  equation  3x=  12,  the  unknown  x  must  have 
the  value  4  on  condition  that  3  a:  shall  represent  12. 

The  members  of  the  conditional  equation  5x4-2  =  6a;— 1  represent  the 
same  number,  17,  only  on  condition  that  the  unknown  x  is  given  the  value  3. 


CONDITIONAL  EQUATIONS  155 

In  the  conditional  equation  x^  —  5x4-6  =  0  the  unknown  x  may  have 
either  of  the  values  2  or  3. 

19.  The  particular  values  which  must  be  given  to  the  unknown 
letters,  in  order  that  both  members  of  a  conditional  equation  may 
represent  the  same  number,  are  called  the  solutions,  or  roots,  of 
the  equation. 

20.  To  solve  a  conditional  equation  is  to  find  its  root,  or  its 
roots  if  it  has  more  than  one,  or  to  show  that  it  has  no  root. 

21.  A  number  or  quantity  is  said  to  satisfy  a  given  conditional 
equation  if,  when  it  is  substituted  for  the  unknown  in  the  equation, 
both  members  may  be  reduced  to  the  same  form,  and  hence  may 
take  the  same  value. 

E.  g.  The  number  5  satisfies  the  conditional  equation  4a;4-l  =6a;  —  9, 
since  when  x  is  replaced  by  5  we  obtain  4x5+l=:6x5  —  9,  or21  =  21, 

22.  Two  conditional  equations  are  said  to  be  eciuivalent,  with 
respect  to  a  specified  unknown,  a;,  when  they  have  the  same  solutions 
with  respect  to  x. 

From  this  definition  it  follows  that  every  solution  of  either  equation 
must  be  a  solution  of  the  other  also ;  that  is,  neither  equation  can 
have  any  solution  which  the  other  has  not. 

E.  g.  The  conditional  equation  3x  —  7  =  5— a;  is  equivalent  to  the  con- 
ditional equation  4  a;  =  12.  Both  equations  are  satisfied  when  x  is  given  the 
value  3,  and  neither  equation  is  satisfied  for  any  other  value  of  x. 

23.  If  the  solution  of  a  given  conditional  equation  cannot  be 
obtained  immediately  by  inspection,  it  is  often  possible  to  derive 
from  it  an  ecjuivalent  conditional  equation  in  which  the  members 
are  of  such  forms  that  the  solution  may  be  readily  obtained. 

E.  g.  It  may  be  seen  that  in  the  conditional  equation  a;  +  1  =  5  the 
unknown  x  must  have  the  value  4;  while  the  fact  that  the  unknown  may 
have  either  of  the  values  1^  or  4^  would  not  appear  immediately  on  inspec- 

/v.  O  2Q  J. 

tion  of  the  conditional  equation =  -7^=-.  < 

^  a;  —  3       27  \ 


166  FIRST  COURSE   IN  ALGEBRA 

General  Principles  governing  Transformations  of 
Conditional  Equations 

24.  Throughout  all  of  the  discussions  which  follow  in  this 
chapter  we  shall  assume  that  none  of  the  functions  considered 
become  infinite  for  any  values  which  may  be  given  to  the  letters 
which  appear  in  them. 

Whenever,  in  this  and  the  following  chapters,  the  word  equation 
is  used,  it  will  be  understood  that  a  conditional  equation  is  meanty 
unless  the  contrary  is  expressly  stated. 

25.  Principle  I.  Substitution. 

If  for  any  ej'presf<ion  in  an  equation  we  substitute  an  identical 
ejcpression,  the  original  and  the  derived  equations  will  be  equivalent. 

E.  g.  Performing  the  imlittated  operations  we  may  reduce  tlie  first  member 
of  the  conditional  equation,  7  a:  +  8  —  2(3  x-\-  A)  =  10  a:  —  5(2  x  —  1),  to 
X,  and  the  second  member  to  5.  That  is,  we  may  replace  the  first  and 
second  membei-s  of  the  given  conditional  ecpiation  by  the  identical  expres- 
sions X  and  5,  and  obtiiin  immediately  the  equivalent  equation  a;  =  5. 

26.  Principle  II.    Addition  and  Subtraction. 

If  identical  ejrpressions  be  added  to  or  subtracted  from  both  mem- 
bers of  an  equation^  the  original  and  derived  equations  will  be 
equivalent. 

This  principle  follows  from  the  Fundamental  Laws  of  Algebra. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

For,  if  A  and  B  represent  expressions   wliich  contain   one  or  several 
unknowns,  and  C  is  either  a  constant  or  any  function  of  the  unknowns,  it 
follows  that  from  A  =  B,  we  may  obtain  either  of  the  equivalent  equations 
A  +  C=B  +  CorA-C=B-C. 

If,  when  certain  values  are  given  to  the  unknowns,  A  and  B  take  the  same 
numerical  value,  that  is,  if  ^  =  B,  then  for  the  same  values  of  the  unknowns 
A  +  C  Avill  take  the  same  value  as  B  +  C,  and  A  —  G  will  take  the  same 
value  lis  B  —  C. 

Hence,  all  of  the  solutions  of  A  =  B  are  solutions  of  both. 
A  +  C=B-]-Cimi\A  -C=B-C. 

Conversely  :  Every  solution  of  either  A-{-C=B-^C  ot  of  A  — C=B  — C 
is  also  a  solution  of  A  =  B. 


EQUIVALENT  EQUATIONS  157 

For,  if  when  certain  values  are  given  to  the  unknowns,  A  -{-  C  and  B  +  C 
have  equal  values,  and  also  A  —  C  and  B  —  C  have  equal  values,  — that  is, 
ifA  +  C=B-{-G  and  A  —  C=  B  —  G,  —  then  for  the  same  values  of  the 
unknowns  we  shall  have  A  +  G—  C=B+C— G  and  also 

A  —  G-\-G=B—G+G;  that  is,  in  either  case  A  =  B. 

Hence,  every  solution  of  either  of  the  derived  equations  A  +  G  =  B  +  G 
or  of  J.  —  G  =  B  —  G  h  also  a  solution  of  the  original  equation  A  =  B. 

Hence,  the  two  equations  are  equivalent,  since  no  additional  solution  is 
gained  by  tlie  transformation. 

27.   From  the  principle  above  we  have  the  following 

Applications  : 

(i.)  Transposition  of  Terms.  Any  term  may  be  stricken  out 
from  either  member  of  an  eqaation  provided  that  a  term  equal  in  ab- 
solute valuSf  but  opjwsite  in  sign,  be  written  in  the  other  member  of 
the  equation. 

This  operation,  which  has  the  effect  of  carrying  a  term  over  from 
one  side  to  the  other  of  the  equality  sign  in  an  equation  and  at  the 
same  time  changing  its  sign  from  +  to  —  or  from  —  to  +,  is  called 
transposition. 

E.  g.    From   the   conditional    equation    4  a:  —  5  =  3  a:  +  2,  we   may,  by 
transposing  the  terms  —  5  and  Sx,  obtain  the  equivalent  e(j[uation 
4a;  —  3a:=2  +  5. 

Combining  the  terms  in  the  members  separately,  we  obtain  the  equivalent 
equation  x—1. 

It  should  be  observed  that,  instead  of  transposing  —  5  from  the  first  mem- 
ber to  the  second  member  of  the  given  equation  4a;  —  5  =  3j:  +  2,  we  may 
add  5  to  both  members  and  obtain  4  a;  —  5  +  5  =  3  a;  +  2  +  5. 

By  combining  the  terms  in  the  first  member  of  this  last  equation,  we 
obtain  the  e<|uiva]ent  equation  4a;  =  3a;  +  2  +  5. 

We  may  cause  the  term  3  a;  to  disappear  from  the  second  member  by 
subtracting  3  a;  from  each  member  of  the  equation.  Hence,  from  the  de- 
rived equation  4  a;  =  3a;  -f  2  +  5,  we  may  obtain  the  equivalent  equation 
4a;  —  3.r=3a;  +  2  +  5  —  3  a;,  which  is  equivalent  to4a;  —  3a;=:24-5. 

This  equation,  as  before,  is  equivalent  to  the  conditional  equation  a;  =  7. 

(ii.)  Identical  terms  may  be  stricken  out  from  both  members  of  an 
equation. 

For,  if  either  of  two  identical  terms  be  transposed  from  one  member  of 
th(;  equation  to  the  other,  then  both  terms  will  appear  with  opposite  signs 


158  FIRST  COURSE  IN  ALGEBRA 

in  the  same  member.     When  combined,  these  terms  will  produce  zero,  and 
accordingly  will  disappear  from  the  derived  equation. 

E.  g.  If  a;  4-  rt  =  5  -f  a,  we  may  immediately  strike  out  a  from  both 
members,  obtaining  as  an  equivalent  equation  a:  =  5. 

(iii.)  From  any  equation  we  may  derive  an  equivalent  equation 
by  reversing  the  sign  of  every  term  in  each  member  from  -{•  to  —  or 
from  —  to  -{-. 

This  operation,  of  reversing  the  signs  of  all  of  the  terms,  has  the 
effect  of  transposing  every  term  in  the  equation  from  each  member 
to  the  other,  and  then  interchanging  the  members  of  the  resulting 
equation. 

E.  g.  Reversing  the  signs  of  all  of  the  terms  of  the  conditional  equation 
—  10x-fl  =  —  9a:  —  2,  we  obtain  the  equivalent  eciuation  +  lOic  —  1  = 
4-  9  X  H-  2.  From  this  equation  we  obtain,  by  transposing  and  combining 
terms,  the  equivalent  equation  x  =  3. 

An  integral  equation  is  said  to  be  in  standard  form  if  the 
second  member  is  zero ;  if  the  first  member  is  reduced  to  simplest 
form  and  arranged  according  to  descending  powers  of  the  un- 
known ;  and  if  the  coefficient  of  the  highest  power  of  the  unknown 
is  positive. 

E.  g.     Each  of  the  following  equations  is  in  standard  form  : 
x^-8x+  12  =  0, 
2aH»  +  3a:2-7a;-t-l=0. 

(iv.)  It  follows  directly  fi-om  the  principle  above  that  any  condi- 
tional equation  may  be  I'educed  to  standard  form  by  transposing  the 
terms  from  the  second  to  the  first  member ^  after  which  the  second 
member  of  the  equivalent  derived  equation  will  be  zero. 

This  operation,  of  transposing  to  the  first  member  every  term 
appearing  in  the  second  member  of  a  given  equation,  has  the  effect 
of  subtracting  the  original  second  member  of  the  equation  from  each 
member  of  the  given  equation,  producing  an  equivalent  equation 
whose  second  member  is  zero. 

E.  g.  From  the  conditional  equation  a:^  =  7  x  —  1 2  we  obtain  the  equiva- 
lent equation  a;^  —  7  a;  -}-  12  =  0,  in  standard  form. 

By  principles  to  be  shown  later,  this  equation  will  be  found  to  have  the 
two  solutions  a;  =  3  and  a:  =  4, 


EQUIVALENT  EQUATIONS  169 

Mental  Exercise  X.     2 

From  each  of  the  following  conditional  equations  derive  an  equiva- 
lent equation  by  transposing  to  the  first  member  the  terms  contain- 
ing X,  and  to  the  second  member  all  other  terms ;  then  simplify  the 
members  separately  : 

1.  a;  -  5  =  0.  32.  a;  -  4  +  c  =  0. 

2.  a;  +  7  =  0.  33.  a;  +  9  +  c?  =  0. 

3.  a;  -  a  =  0.  34.  2  a;  =  13  -f-  a;. 

4.  x^rh  =  0.  35.  4a;  =  3a;  +  5. 

5.  0  =  9  —  X.  36.  6a;  =  8  4-  5a;. 

6.  0  =  12  -  a;.  37.  9a;  =  8a;  —  15. 

7.  0  =  -  a;  +  14.  38.  23  a;  =  22  a;  -  24. 

8.  0  =  -a;— 17.  39.  3a;  — 25  =  2a;. 

9.  a;  -2  =  1.  40.  12  a;  -  7  =  11  a;. 

10.  a;  —  8  =  2.  41.   15  a;  +  4  =  14  x. 

11.  a;  —  9  =  13.  42.  —  3  a;  —  10  =  —  4  a;. 

12.  a;+  3  =  7.  43.   17  -f  30a;  =  29a;. 

13.  a;  +  7  =  8.  44.  14  —  8  a;  =  —  9  a;. 

14.  6 -Fa;  =11.  45.  2  a; -f  5  =  a;  +  7. 

15.  8  -Ha;  =  14.  46.  3  a;  -  1  =  2a;  4-  6. 

16.  a; -F  9  =  2.  47.  4a;  —  7  =  3a;  —  2. 

17.  a;  +  11  =  12.  48.  5a;  -}-  8  =  4a;  -h  5. 

18.  a;  +  5  =  5.  49.   7  a;  -f  3  =  6  a;  —  5. 

19.  a;- 4  =  4.  50.  10a;  +  11  =  9a;-F  7. 

20.  a;  -I-  7  =  —  7.  51.  13  a;  —  6  =  12  a;  —  5. 

21.  3  =  4  — a;.  52.  6  a; -f  17  =  23  +  5  a;. 

22.  4  =  9  — a;.  53.  15  a;  —  1  =  11  -f  14  a;. 

23.  5  =  11  —  a;.  54.  13  -[-  12  a;  =  11a; -|-  21. 

24.  6  =  4  — a;.  55.  4  —  15  a;  =  11  —  16  a;. 

25.  12  =  5  —  a;.  56.  16  —  17  a;  =  —  18  a;  —  19. 

26.  8  =  -  3  —  a;.  57.  5  —  19a;  =  —  6  —  20a;. 

27.  -  15  =  11  -  a;.  58.  —  19  a;  —  33  =  —  20a;  —  31. 

28.  -  5  =  -  a;  -  2.  59.  -  28  a;  -  37  =  -  40  -  29  x. 

29.  —  12  =  -a;—  18.  60.  —  31  a;  —  11  =  29  —  32  a;. 

30.  a;— 1— a  =  0.  61.  21  a; -f- ^  =  1 -f  20a;. 

31.  a; -f  2-^^  =  0.  62.  26a; -f-  i  =  25a;-f  1. 


160  FIRST  COURSE  IN  ALGEBRA 

63.  27  35  +  i  =  2  4-  2G35.  74.  ^^x  +  18  =  -  tV«  -  18. 

64.  2335  —  i  =  22  3^+1.  75.  ^' x  —  23  =  23  +  ^35. 

65.  3435  +  f  =  333-  +  1.  76.  3535+  1  =  «  +  3435. 

66.  |35  +  1  =  i 35  +  G.  77.  37 35  —  2  =  36 35  +  6. 

67.  |35  +  2  =  |35+  9.  78.  4035  +  c  =  3935+  5. 

68.  §  35  —  7  =  9  +  ^  35.  79.  42  35  —  c^  =  41  35  -  6. 

69.  f  35—  13  =  4  — ^35.  80.  m- 4835  =  4  -4935. 

70.  §35- 10  =  3- f  35.  81.  5035+ i  =  4935  +  a. 

71.  |35-|  =  f-|3-.  82.    5435  +  ^>  =  i  +  5335. 

72.  A  a;  —  i  =  5  -  i\  a5.  83.  61  35  +  ^>  =  a  +  GO35. 

73.  hx-\^  =  \\-hx.  84.  6335- c  =  62  35  + 6. 

From  each  of  the  following  conditional  ec^uations  derive  an 
equivalent  equation  by  omitting  the  identical  terms  from  both 
members  : 

85.  35  +  ^  =  ^  +  c.  90.  a35  +  3;  =  «35  +  2. 

86.  35  +  /j  =  ff  +  ^>.  91.  ^w  +  35  =  3  +  hx. 

87.  X  —  d  =  b  —  d.  92.  mx  —  1  =  35  +  mx. 

88.  m  •\-  d  =  m  +  35.  93.  35  +  a  =  <7. 

89.  «  —  ^  =  35  —  6.  ^X.  X  —  cx  =  —  d  —  ex. 

From  each  of  the  following  conditional  equations  derive  an  equiv- 
alent equation  in  the  standard  form  J  =  0  : 

105.  63;2  — 435=  535^+  21. 

106.  235^  +  435  =  33;+  3. 

107.  8  35^+35=735—  1. 

108.  7352+73;=  10  — 535^. 

109.  33;'+  93;+  6  =  435+  8. 

110.  435'-  13  =  335^+  5  —  335. 

111.  35^-1235+4=1035— 7  3;2—ll. 

112.  935'+G3;  =  635+3G  +  83;2 

113.  535  —  9  =  3  —  3352+53;. 

114.  35^+53;2+935+7=3;^+43;2— 11. 
+  3;  +  6  =  3.^  —  4  35'^  +  11  3;  —  18. 

4  .^  —  5  =  35*  +  G  3;=^  —  5  35  +  7. 


95. 

35^  —  a;  =  12. 

96. 

35^— 5  =  4ar. 

97. 

35^+  635=16. 

98. 

35^=735-30. 

99. 

35^=22-935. 

100. 

05^=  10  35+  39. 

101. 

235^=3-535. 

102. 

33;2_  5  =  14^ 

103. 

2335  =  6-4352. 

104. 

10352=1335+  3. 

115.    35»-  3  352H 

116.    35*+   7352- 

EQUIVALENT  EQUATIONS  161 

28.   Principle  III.      Multiplication 

If  both  member's  of  an  integral  equation  be  multiplied  by  the  same 
number  or  expression^  the  original  and  the,  derived  equations  will  be 
equivalent^  provided  that  the  multiplier  used  is  neither  zero  nor 
infinitely  great,  and  that  it  does  not  contain  the  unknown  letter  or 
letters. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  a  conditional  equation  be  represented  by  A  —  B  (1),  in  which  either 
A  or  i>,  or  both,  are  functions  of  some  unknown  letter,  x. 

If  G  19,  Q.  number  which  is  neither  zero  nor  infinitely  great,  or  if  G 
represents  an  expression  which  does  not  contain  the  unknown  letter  x, 
then  the  given  equation  ^  =  ^  is  equivalent  to  the  derived  equation 
AG  =  BG{^y 

To  show  this  we  will  write  the  derived  equation  (2)  in  the  standard 
form  ^(7-5(7  =  0(3). 

Since  G  is  assumed  to  be  neither  zero  nor  infinitely  great,  we  may  divide  the 
expression  AG  —  BG  hy  G  and  write  equation  (3)  in  the  form  G  (A  —  B)  = 
0  (4). 

Every  value  of  x  which  satisfies  the  original  equation  A  =  B  reduces  the 
factor  A  —  B  to  zero. 

Since  G  is  assumed  to  have  a  value  which  is  not  infinitely  great,  it  follows 
that  any  value  which,  when  substituted  for  x,  reduces  the  factor  A  —  B  to 
zero,  reduces  the  product  G(A  —  B)  to  zero.  Accordingly  such  a  value  of  a: 
satisfies  the  equation  G(A  —  B)=  0. 

It  follows  that  every  solution  of  (I)  is  also  a  solution  of  (3),  and  hence 
o(AG=BG(2). 

That  is,  no  solution  of  (I)  is  lost  hy  the  transformation. 

To  show  that  equation  (1)  is  equivalent  to  equation  (2)  it  remains  for  us 
to  show  that  no  solutions  have  been  gained  in  passing  from  equation  (1)  to 
equation  (2). 

Every  value  of  x  which  satisfies  G(A  —  B)  =  0  (4)  reduces  the  product 
G(A  -  B)  to  zero. 

In  order  that  the  product  of  two  factors  shall  become  zero  it  is  necessary 
and  sufficient  that  one  of  these  factors  shall  become  zero,  the  other  factor 
not  becoming  infinitely  great. 

Accordingly,  since  the  value  of  G  is  assumed  to  be  neither  zero  nor  infi- 
nitely great,  it  is  necessary  that  the  remaining  factor  A  —  B  should  become 
zero. 

Every  value  which,  when  substituted  for  x,  reduces  A  —  B  to  zero, 
must  satisfy  the  equation  A  —  B  =  0.     It  follows  that  every  solution  of 

11 


162  FIRST  COURSE  IN  ALGEBRA 

C(A  —  B)  z=0i3  a  solution  also  of  A  —  B  =  0,  and  hence  is  a  solution  of 
A  =  Bil). 

That  is,  no  solution  is  gained  by  the  transformation. 

Since,  by  the  reasoning  above,  solutions  are  neither  gained  nor  lost  in 
passing  from  the  given  equation  A  =  B  to  the  derived  equation  AC  =  BG, 
these  equations  are  equivalent. 

E.  g.  From  the  equation  a:  +  1  =  f  x  +  ^,  we  may,  by  multiplying  both 
members  by  the  constant  multiplier  3,  derive  the  equivalent  equation 
3x  +  3  =  2a:  +  7. 

Transposing  and  combining  terms,  we  obtain  from  this  last  equation  x  =  4. 

Since  the  equation  x  =  4  has  the  single  solution  4,  it  follows  that  the 
given  equation  to  which  it  is  equivalent  must  have  the  solution  x  z=z  4,  and 
no  other. 

29.  Caution.  To  be  certain  that  solutions  have  not  been  gained 
in  solving  conditional  ecjuations,  it  is  necessary  that  we  know  that 
the  multipliers  with  which  we  affect  the  forms  of  given  equations 
are  different  from  zero. 

30.  The  solutions  which  may  be  gained  when  the  members  of  a 
con"ditional  equation  are  transformed  by  multiplication  may  be 
determined  by  the  following 

Principle  Relating^  to  Extra  Roots :  77is  strange  or  extra 
solutions  which  may  be  introduced  into  the  members  of  a  derived 
equation  by  multiplying  the  members  of  a  given  equaticm  by  an  integral 
expression  containing  the  unknown  number,  are  the  values  of  the  un- 
known which,  when  substituted,  reduce  the  multiplier  to  zero. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

If  C  were  a  function  of  x  it  might  happen  that,  when  particular  values 
were  given  to  x,  C  might  l)ecome  zeio  and  A  —  B  become  different  from 
zero. 

Such  values  of  x  would  satisfy  the  equation  C  {A  —  B)  =  0  without  re- 
ducing the  fjictor  ^  —  ^  to  zero,  and  hence  without  satisfying  the  given 
equation  ^  =  jB  (1)  §  28. 

Thus,  if  C  were  a  function  of  a;,  solutions  of  the  derived  equation  AG=  BG 
might  exist  which  were  not  solutions  also  of  the  original  equation  A  =  B. 
In  such  a  case  the  given  and  derived  equations  would  not  be  equivalent. 

E.  g.  If  both  members  of  the  equation  3ar—  16  =  a:  —  2  (1)  were  multi- 
plied by  the  expression  a:  —  3,  containing  the  unknown,  we  should  obtain  the 
equation  (3  a;  -  16) (a;  _  3)  =  (x  -  2)(a:  -  3),  (2). 


EQUIVALENT  EQUATIONS  .163 

The  derived  equation  (2)  would  have,  not  only  the  solution  a;  =:  7  of  the 
original  equation  (1),  but  also  the  solution  x  =  3  which  does  not  satisfy  the 
original  equation  (1),  This  extra  solution  x  =  3  would  have  been  intro- 
duced by  means  of  the  multiplier  x  —  '3  which  becomes  zero  for  a?  =  3. 

By  using  the  multiplier  a:  —  3  we  should  thus  gain  a  solution  in  passing 
from  the  given  equation  (1)  to  the  derived  equation  (2). 

31.  Principle  IV.  Division. 

1/  both  members  of  an  equation  be  divided  by  the  same  number^  the 
original  and  the  derived  equations  will  be  equivalent^  provided  that 
the  divisor  used  is  neither  zero  nor  infinitely  great^  and  that  it  does 
not  contain  the  unknown  letter  or  letters. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Instead  of  division  by  a  number  D  we  may  substitute  multiplication  by 
its  reciprocal  y-. 

Hence,  since  D  can  become  neither  zero  nor  infinitely  great,  it  follows 
that  1  jD  can  become  neither  infinitely  great  nor  zero. 

Accordingly,  the  principle  under  consideration  is  proved  by  the  course 
of  reasoning  employed  for  the  proof  of  Principle  III  §  28,  provided  that 
1/7)  is  represented  in  that  proof  by  G. 

E.  g.  From  the  equation  2  a:  —  4  =  10,  we  shall,  by  dividing  both  mem- 
bers by  the  constant  2,  derive  the  equivalent  equation  a:  —  2  =  5.  Trans- 
posing and  combining  terms,  we  have  a;  =  7. 

Since  the  given  and  derived  equations  are  equivalent,  it  follows  that  7, 
which  is  the  single  solution  of  the  last  equation,  must  be  the  single  solution 
of  the  given  equation. 

32.  Caution.  Beginners  often  make  the  error  of  dividing  both 
members  of  an  equation  by  an  expression  containing  the  unknown, 
and  thus  lose  as  solutions  of  the  given  equation  such  values  as 
would,  when  substituted  for  the  unknown,  reduce  to  zero  the 
divisor  thus  used  and  rejected. 

33.  Principle  Relating-  to  Loss  of  Roots :  If  identical  ex- 
pressions containing  the  unknown  be  removed  by  division  from  both 
members  of  an  equation^  the  given  and  the  derived  equations  will  not 
be  equivalent. 

The  solutions  of  the  given  equation  which  are  not  solutions  of  the 
derived  equation  also  are  those  values  of  the  unknown  which,  if  sub- 
stitutedj  would  reduce  to  zero  the  expression  removed  by  division. 


164  FIRST  COURSE  IN  ALGEBRA 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

It  follows  from  the  reasoning  employed  in  the  proof  of  the  principle  of 
§  30  that  the  solutions  of  the  equation  AG  =  BG  are  the  same  as  those  of 
the  two  equations  A  =  B  and  G  =  0,  provided  that  A,  B,  and  6^ are  functions 
of  the  unknown. 

Hence,  it  follows  that  if  (7,  which  is  common  to  both  members  of  the 
et^uation  AG  =  BG,  be  removed  by  division  and  rejected,  the  solutions  of 
the  equation  formed  by  placing  this  divisor  equal  to  zero,  that  is  C  =  0, 
are  thus  lost. 

E.  g.    If  we  divide  both  members  of  the  conditional  equation 
(a:  -  l)(x  -  4)  =  2(x  -  1) 
by  the  divisor  x  ~l  containing  the  unknown  x,  we  shall  obtain  as  a  derived 
equation  x  —  4  =  2.     This  derived  e^j^uation,  a;  —  4  =  2,  is  satisfied  by  the 
single  value  a:  =  6.     We  have,  by  the  process,  lost  a  root,  since  the  original 
etjuation  is  satisfied  by  the  value  a;  =  1  in  addition  to  the  value  x  =  6. 

It  should  l3e  observed  that  the  expression  a;  —  1,  which  was  removed  from 
both  members  of  the  given  equation  by  division,  becomes  zero  when  x  is 
given  the  value  1,  which  is  the  root  which  was  lost  when  passing  from  the 
given  to  the  derived  equation. 

The  conditional  equation  a:*  =  6  a;  is  satisfied  by  the  values  x  =  0  and 
X  =  6,  and  by  no  others.  If,  however,  we  derive  the  equation  a:  =  6  by 
removing  by  division  and  rejecting  x  from  both  members  of  a:^  =  6  a:,  we 
shall  lose  one  of  the  solutions  of  the  original  equation,  namely  x  =  0. 

34.   From  the  principles  of  §§  28,  31  we  have  the  following 

Applications  : 

(i.)  If  the  terms  of  an  equation  are  integral  with  reference  to  some 
specified  letter,  x,  and  the  coefficients  of  x  are  fractional,  we  may 
derive  an  equivalent  equation  in  which  the  coefficients  of  x  are  inte- 
gral, hy  multiplying  all  of  the  terms  of  the  given  equation  hy  the  hast 
number  which  contains  all  of  the  denominators  of  the  fractional 
coefficients  exactly  as  divisors. 

E.  g.  If  we  multi])ly  the  terms  of  both  members  of  the  conditional 
equation  6x  —  106  =  |a;  +  J^  a;  +  100  by  20,  which  is  the  least  number 
which  can  be  divided  without  remainder  by  all  of  the  denominators  of  the 
fractional  coefficients,  we  shall  obtain  the  equivalent  equation,  120  a:  —  2120 
==  15  a;  +  2  X  +  2000,  in  which  the  coefficients  are  integral. 

Transposing  and  collecting  terms  in  the  derived  equation,  we  obtain 
103  a:  =  4120,  the  solution  of  which  is  found  to  be  a;  =  40. 


EQUIVALENT   EQUATIONS  165 

(ii.)  From  an  integral  equation  may  be  derived  an  equivalent 
equation  in  which  the  coefficient  of  any  specified  term  shall  have  any 
desired  value. 

It  is  often  desirable  so  to  transform  the  terms  of  an  equation  that 
the  coefficient  of  the  highest  power  of  the  unknown  shall  be  unity. 

E.  g.  By  dividing  each  term  of  the  conditional  equation  Sa:^  +  4  a:  =  7 
by  3,  we  may  derive  the  equivalent  equation  x^  -\-  ^x  =  \,  in  which  the 
coeflScient  of  the  highest  power  of  x  is  unity. 

35.  When  by  means  of  any  step  we  derive  an  equation  which  is 
equivalent  to  another,  this  step  is  said  to  be  reversible,  since, 
either  equation  being  equivalent  to  the  other,  we  may  derive  either 
equation  from  the  other,  and  hence  take  the  step  forward  or  back- 
ward without  gaining  or  losing  solutions. 

36.  Observe  that  we  may  derive  an  equivalent  equation  by  adding 
to  or  subtracting  from  both  members  of  a  given  equation  the  same 
expression  containing  the  unknown.  But  in  case  we  multiply  or 
divide  both  members  by  an  expression  containing  the  unknown,  the 
resulting  equation  may  or  may  not  be  equivalent  to  the  one  from 
which  it  is  derived,  because  in  certain  cases  we  may  gain  or  lose 
solutions. 

Mental  Exercise  X.     3 

From  each  of  the  following  conditional  equations  derive  an  equiva- 
lent equation  in  which  the  coefficient  of  x  is  unity : 

1.  2iK  =  8.  14.  9a;  =  —  54.  27.  5cc  =  2. 

2.  3 a;  =15.  15.  lla^  =  — 77.  28.  lx='d. 

3.  4a;  =12.  16.  —16  =  8a;.  29.  8a;  =  5. 

4.  5a;  =30.  17.  -7  =  7a;.  30.  9a;  =  2. 

5.  6a;  =  42.  18.  -22  =  2a;.  31.  3  =  11  a;. 

6.  7a;  =  56.  19.  3a;=l.  32.  4=13a;. 

7.  9a;  =  81.  20.  4a;=l.  33.  —5  =  14a;. 

8.  14  =  2  a;.  21.  5a;  =  — 1.  34.  3  a;  =  4. 

9.  18  =  3a;.  22.  l  =  6a;.  35.  2a;  =  5. 

10.  25  =  5a;.  23.  1  =  8a;.  36.  4a;  =  7. 

11.  33  =  lla;.  24.  —  l  =  9a;.  37.  5a;  =13. 

12.  42  =  14a;.  25.  2a;  =  0.  38.  6a;  =19. 

13.  8 a;  =  —  24.  26.  0  =  3a;.  39.  7 a;  =  29. 


166  FIRST  COURSE   IN  ALGEBRA 

40.  8  a;  =  43.  77.  4i«  =  ff  114.  f£c=l. 

41.  9a;  =  —  28.  78.  5a;  =  Y.  115.  fa;=l. 

42.  12  a;  =  -25.  79.  ia;  =  2.  116.  f£c=l. 

43.  24  =  7a;.  80.  ia;  =  3.  117.  i^aj=l. 

44.  32  =  9«.  81.  ia;  =  5.  118.  §a5  =  0. 

45.  43  =  11  a;.  82.  ia;  =  4.  119.  l=J^a;. 

46.  —49  =  13a;.  83.  }a;  =  9.  120.  1  =  |a;. 

47.  -52  =  12a;.  84.^^^05=1.  121.  1=^^. 

48.  6a;  =  3.  85.  T»5a;=12.  122.  |a;  =  2. 

49.  8a;  =  2.  86.  ^x  =  —  10.  123.  ^■^^x  =  4. 

50.  18 a;  =  6.  87.  TVa;  =  — 3.  124.  4a;  =  3. 

51.  42a;=7.  88.  i^ja;  =  —  13.  12^.  ta;  =  5. 

52.  65a;=5.  89.  ia;  =  0.  126.  fa;  =  4. 

53.  34a;  =  2.  90.  6  =  |a;.  127.  T%a;  =  8. 

54.  50a;  =  — 10.  91.  12  =  ia;.  128.  ^a;  =  9. 

55.  3a;  =  a.  92.  — 14  =  Ja;.  129.  ^a;  =  6. 

56.  4a;  =  6.  93.  ^x  =  ^.  130.  Ja;=7. 

57.  5a;  =  c.  94.  ^a;=i.  131.  fa;  =  6. 

58.  6a;  =  c?.  95.  ia;  =  |.  132.  |a;  =  f 

59.  7a;  =  27W.  96.  ^a;  =  i.  133.  fa;=f 

60.  8a;=3w.  97.  ^x  =  ^.  l'34.  ^x  =  ^j. 

61.  9a;=5A:.  98.  i^y a;  =  ,^5.  135.  |a;=f. 

62.  6rt=lla;.  dd.  ^x  =  j^.  136.  ^x  =  %. 

63.  126=17a;.  100.  5^5  a;  =  ^xr-  137.  fa;  =  §. 

64.  10a;  =  8a.  01.  ia;  =  i.  138.  |a;  =  f 

65.  12a;=146.  102.  |a;  =  i  lSd.^  =  ^x. 

66.  16  a;  =  18  c.  103.  J  a;  =  f  140.  \^  =  ^x. 

67.  21c?=14a;.  104.  ^x  =  i.  141.  ^jy  =  ^x. 

68.  33^  =  6a;.  105.    ix  =  ^.  142.  f^  =  |a;. 

69.  3a;  =  i  106.  ^-x  =  i.  143.  \x  =  ^a. 

70.  4a;  =  i.  107.  ^V^^^i-  144.  ia;  =  i6. 

71.  5a;  =  f  108.  i^x  =  \.  145.  ^x  =  \c. 

72.  6a;  =  i.  109.  ^  =  ^a;.  146.  \x  =  ^d. 

73.  8a;  =  i  110.  \  = -i^x,  147.  ia;  =  ia. 

74.  9a;  =  — i.  111.  i  =  ia;.  148.  ^x  =  \h. 
lo.2x  =  ^.  112.  fa;  =1.  Ud.^x=^m. 
16.  Sx  =  \\  113.  |a;=l.  150.^x  =  ^a 


EQUIVALENT  EQUATIONS  167 


151. 

^hx  =  ijb. 

163. 

^i  h  =  tV  X. 

175. 

^x  =  }a. 

152. 

\x  =  ^a. 

164. 

i^k^i^x. 

176. 

lx  =  lk 

153. 

^x=^b. 

165. 

■s^m=^^^x. 

177. 

^x  =  ic. 

154. 

^l^X  =  j\jC. 

166. 

^§a  =  ^jx. 

178. 

t  ^  =  f  «. 

155. 

^x  =  ^d. 

167. 

^ja  =  ^jsx. 

179. 

^X  =  ^jb. 

156. 

^x  =  j\h. 

168. 

aV^^^VaJ- 

180. 

tic  =  fc. 

157. 

^X=j\k. 

169. 

^x=  ^a. 

181. 

%x  =  ^d. 

158. 

\a  =  ^x. 

170. 

^x  =  ^b. 

182. 

lx  =  ^m. 

159. 

^b  =  \x. 

171. 

\x  =  %c. 

183. 

^a  =  ^x. 

160. 

^c  =  \x. 

172. 

hx  =  ^d. 

184. 

rb  =  'zx. 

161. 

id=^,x. 

173. 

^m  =  ^x. 

185. 

^c  =  ^x. 

162. 

l9  =  ^x. 

174. 

^m  =  ^x. 

186. 

id  =  ^x. 

168  FIRST  COURSE  IN  ALGEBRA 


CHAPTER  XI 

EQUATIONS  OF  THE  FIRST  DEGREE  CONTAINING  ONE 
UNKNOWN 

1.  A  simple  or  linear  equation  containing  a  single  unknown 
is  an  equation  which  is  of  the  first  degree  with  reference  to  the 
unknown  appearing  in  it. 

2.  Solution  of  a  Linear  Equation.  By  applying  the  Prin- 
ciples of  Chapter  X.,  any  integral  rational  equation  containing  one 
unknown,  that  is  any  linear  equation,  may  be  transformed  into  an 
equivalent  equation  of  the  standard  form 

ax  +  b  =  0,  (1) 

in  which  a  and  h  are  either  simple  or  compound  expressions,  both 
free  from  the  unknown.     The  term  6,  which  is  free  from  the  un- 
known, may  be  zero,  but  the  coefficient  a  cannot  be  zero. 
Equation  (1)  may  be  written  in  the  equivalent  form 

ax  =  —  h.  (2) 

Since  a  :?^  0,  we  may  divide  both  members  by  a,  and  obtain 

.  =  ^.  (3) 
It  follows  that  any  linear  equation  in  the  form  ax  -\-h  —  0,  con- 
taining one  unknown,  has  one  solution  and  one  only,  x  = . 

Ex.  1.    Solve  the  equation  15  a:  —  11  =  5  a:  +  9.  (1) 

Transposing  the  terms  5  x  and  —  1 1  we  obtain 

15a;-5a:=9-f  11,  (2) 

which  is  equivalent  to  equation  (1)  by  Chapter  X.  §  27  (i.)  Principle  II. 
Combining  like  terms,  we  obtain 

10  a;  =20,  (3) 

which  is  equivalent  to  equation  (2)  by  Chapter  X,  §  25  Principle  I. 


LINEAK  EQUATIONS  169 

Dividing  both  members  by  the  coefficient  of  x,  we  obtain  the  solution 

a;  =  2. 

Since  throughout  the  entire  process  all  of  the  equations  obtained  are 
equivalent  to  one  another,  the  solution  of  the  last,  a:  =  2,  is  a  solution  of, 
and  the  only  solution  of,  the  first. 

We  may  verify  the  solution  by  substituting  this  value,  x  =  2,  in  the 
original  equation  as  follows  : 

15 -2- 11  =  5-2  +  9 
19=  19. 
Ex.  2.    Solve  the  equation  12  -  7  a;  =  a;  -  18.  (1) 

Transposing  the  terms  so  that  all  terms  containing  x  shall  appear  in  the 
first  member,  and  all  terms  free  from  x  in  the  second,  we  have  the  equiva- 
lent equation 

-7a; -a;  =  -18 -12.  (2) 

Combining  terms,  we  derive  the  equivalent  equation 

-  8  a;  =  -  30.  (3) 

Dividing  both  members  by  the  coefficient,  —  8,  of  x,  we  obtain  as  a 
solution  of  this  last  equation 

_-30_  15 
^~  -8  ~  4' 
Since,  in  obtaining  the  successive  equations  we  have  applied  the  prin- 

15 
ciples  f(jr  deriving  equivalent  equations,  the  result  — ,  which  is  the  only 

solution  of  the  last  equation,  must  be  a  solution,  and  the  only  solution  of 
the  given  equation. 

Substituting  this  value  in  the  original  equation,  we  obtain 
12  -  7  (V)  =  J^  -  18 
-I4i  =  -14i. 

Ex.  3.   Solve  the  equation  |(a;  -  5)  +  f  (x  -  3)  =  a;  -  |.  (1) 

Multiplying  both  members  of  the  equation  by  15,  which  is  the  least 

number  which  contains  the  denominators  of  the  different  fractions  exactly 

as  divisors,  we  obtain  the  equivalent  equation 

3  (a;  -  5)  +  5  •  2  (a;  -  3)  =  15  a;  -  3  .  7.  (2) 

Performing  the  indicated  operations,  we  obtain 

3a; -15  + 10a; -30=  15a; -21, 
and  hence  13  a;  -  45  =  15  a;  -  21.  (3) 

By  Principle  I,  Chapter  X,  §  25,  this  equation  is  equivalent  to  equation 
(2)  above. 


170  FIRST  COURSE  IN  ALGEBRA 

Transposing  terms,  we  derive  from  (3)  the  equivalent 
equation  13  a:  —  15  a;  =  —  21  +  45.  (4) 

From  (4)  we  obtain,  —  2  a;  =  +  24.  (5) 

Dividing  both  members  by  the  coefficient,  —2,  of  ar,  we  obtain  the  solii- 

24 
tion  X  =  — -  =  —  12. 

—  ji 

Since  all  of  the  steps  used  in  the  process  of  deriving  these  equations  suc- 
cessively are  reversible,  the  solution  of  the  last  equation  must  be  a  solution 
of,  and  the  only  solution  of,  the  original  equation. 

Verifying  the  accuracy  of  the  result  by  substituting  in  the  first  equation, 
we  have 

i  (-  12  -  5)  +  H-  12  -  3)  =  -  12  -  ^  ' 
-  13f  =  -13f 

3.  It  is  possible  that  roots  may  be  either  gained  or  lost  during 
the  process  of  solution  of  a  conditional  equation.  (See  Chapter  X. 
§§  30,  33). 

The  substitution  in  the  original  equation  of  any  roots  which  may 
satisfy  any  one  of  the  derived  equations  does  not  establish  thereby 
the  equivalence  of  the  equations. 

By  such  a  substitution  we  simply  determine  whether  or  not  the 
values  substituted  are  solutions  of  the  original  equation. 

The  equivalence  of  the  given  and  derived  equations  must  be 
determined  by  examining  the  process  of  derivation. 

Ex.  4.    Solve  the  equation  (a:  +  2)(x  +  3)  +  4  =  (a;  +  2)2.  (1) 

Performing  the  indicated  multiplications,  we  obtain 

a:2  +  5  a;  +  6  +  4  =  a;2  +  4a;  +  4.  (2) 

Equal  terms,  such  as  x^  or  4,  which  occur  in  both  members,  may  be 
stricken  out,  since  if  transposed  they  would  by  combination  produce  zero. 

Hence  we  have  5a;  —  4x  =  —  6  (3) 

a;  =  -  6. 

No  root  is  gained  or  lost  by  any  of  the  operations  performed  on  the 
members  of  these  equations.  Hence,  the  root  of  the  last  equation  is  a 
solution  of,  and  the  only  solution  of,  the  original  equation. 

Verifying  the  result  by  substitution,  we  obtain 

(-  6  +  2)(-  6  +  3)  +  4  =  (-  6  +  2)2 
16  =  16. 

4.  The  ultimate  test  of  the  correctness  of  every  solution  is  that, 
when  the  value  found  is  substituted  for  the  letter  representing  it, 


LINEAR  EQUATIONS  171 

the  equation  is  satisfied.  No  matter  how  we  may  obtain  the  value 
of  an  unknown  letter,  even  if  it  be  by  mere  guessing  or  by  inspec- 
tion, if  it  stands  this  test  of  substitution,  it  is  a  solution. 

5.  The  process  of  solving  a  conditional  equation  consists  of 
obtaining  and  substituting  for  a  given  equation  another  equation 
which  has  all  of  the  solutions  of  the  first,  and,  if  possible,  no  more 
solutions,  and  which  is  of  such  form  that  the  relations  expressed 
between  the  letter  whose  value  is  to  be  found  and  the  remaining 
quantities  in  the  equation  is  less  complicated.  This  process  is 
continued  until,  if  possible,  an  equation  is  finally  obtained  which 
may  be  solved  by  inspection. 

6.  Suggestions  Concerning  the  Solution  of  Simple  Equa- 
tions Containing  One  Unknown  Number 

(i.)  Remove  firactional  coefficients,  if  there  be  any,  by  multiply- 
ing both  members  of  the  equation  by  the  least  number  which  con- 
tains the  denominators  of  the  different  fractions  exactly  as  divisors. 

(ii.)    Perform  all  such  indicated  operations  as  are  necessary  to 
separate  the  terms  of  the  equation  into  two  distinct  groups, — one. 
group  consisting  of  all  of  the  terms  containing  the  unknown  num- 
ber (and   no  other  terms),  and  a  second  group  consisting  of  all 
terms  which  do  not  contain  the  unknown  number. 

The  terms  which  contain  the  unknown  number  are  commonly 
transposed  to  the  first  member  of  the  derived  equation,  and  all 
terms  which  are  fi:ee  from  the  unknown  are  transposed  to  the 
second  member  of  the  equation. 

(iii.)  All  numerical  or  all  monomial  or  pol3momial  factors  not 
containing  the  unknown  numbers,  which  are  common  to  all  of  the 
terms  of  both  members  of  the  equation,  should  be  removed  by 
division  and  rejected  as  soon  as  discovered. 

(iv.)  Combine  into  one  term  all  of  the  terms  containing  the 
unknown  number,  and  into  another  term  the  remaining  terms 
which  are  free  from  the  unknown. 

(v.)  Divide  both  members  of  the  equation  by  the  coefficient  of 
the  unknown  number. 

(vi.)  The  expression  found  for  the  unknown  number  should  be 
reduced  to  simplest  form. 


172  FIRST  COURSE  IN  ALGEBRA 

Exercise  XI.     1     (Mental  and  Written  Examples) 

Solve  the  following  equations,  verifying  all  results  by  substitution. 
The  first  sixty-four  examples  may  be  solved  mentally : 

1.  4ic-f  5  =  3ic4-  7.  24.  2a;  — 37  =  lx+  3. 

2.  6«+  11  =  5ic+  17.  25.  Sx+  5  =  dx-\-  59. 

3.  10a-  7  =  9a;  4-  8.  26.  5a;+  1  +  2.'k  =  15. 

4.  7a;—  10-6a;  =  0.  27.  6a;- 2  +  3iK  =  25. 

5.  5a;  +  11  -4a;  =  0.  28.  7a;  +  10 -4a;  =  40. 

6.  11a;- 6- 9a;  =  0.  29.  9a;  —  2  —  4a;  =  —  57. 

7.  3a;+ 1  =a;— 1.  30.  12a;  =  8  +  30  -  a;. 

8.  9a;+l  =  3a;+5.  31.  5a;-4+6a;-7  =  0. 

9.  13a;+ 4=  lla;+  10.  32.  13  a;  —  50  —  2a;  =  a;. 

10.  6a;—  11  =  2a;  4-  9.  33.  19a;— 5  +  2a;  =  39. 

11.  13a;- 18  =  a; -h  6.  34.   18a;  —  33  —  7a;  =  3a;  -  1. 

12.  8a;— 13  =  3aj-53.  35.  4a;  +  5  +  6a;  =  7  +  8a;+ 4. 

13.  22a;  +  15  =  19a;— 12.  36.  5a;  +  7  +  3a;  =  a;+ 10  +  Ga;. 

14.  15a;+  37  =  3a;+  13.  37.  8a;+ 3  4- 6a;  =  4a;+ 11  +  9a;. 
.    15.  12a;+  1  =  28  +  3a;.  38.  9a;— 7-5a;=  11  a;+ 5  — 8a;. 

16.  19  +  17a;  =  59  — 3a%  39.  9a;— 7  — 4a;=  10a;  + 5  — 7a;. 

17.  15a;— 13  =  29  + 8a;.  40.  15  —  7  a;+ 6  =  12- 2a;+ 19. 

18.  21  +  22 a;  =  8 a;  — 35.  41.  5a;  +  12— 8a;=  19  — 13a;  +  2. 

19.  7a;+2  =  4a;+7.  42.  22a;— 9  —  6a;  =  5a;  — 6  +  a;. 

20.  17  — 18a;  =  87  — 25a;.  43.  7a;  —  5  +  2a;  =  3a;  +  8  +  a;. 

21.  15  +  11a;  =  79  — 5a;.  44.  4a;+ 9- 7a>=  8a!— 11  +  3a;. 

22.  2a;+ 23  =  5a;+ 2.  45.   ll  +  7a;  — 18-3a;=9  +  a;+5. 

23.  4a;  — 23  =  1  — 4a%  46.   12+ 11  a;+ 3  =  5a;  — 2  — 4a;. 

47.  19a;+ 9— 12a;+ 6  =  2a;  + 35  + 3a;. 

48.  25+  12a;— 23  +  14a;  =  25a;+  12-23a;+ 2. 

49.  8  — 4a;-2  + 9a;=  7  +  2a;- 19- 6a;. 

50.  7a;+ 9  — 3a;  + 5  =  4a;— 11  +  2a;  +  45. 

51.  3a;+  13  +  5a;- 7  =a;+ 7  +  2a;+ 2. 

52.  14a;— 1  +  3a;+ 5  =  7a;  + 2  — 4a;  — 8. 

53.  5a;—  (2a;  +  3)  =  12. 

54.  3a;—  (13  —  a;)  =  61. 

55.  12-  (4  a;  +  7)  =  13. 

56.  17  a;  +  5(2- 3  a;)  =  18. 


LINEAR  EQUATIONS  ,   1T3 

57.  9a5-2(l  +  4a?)  =  3. 

58.  17-3(aj+  11)  =  - 7a;. 

59.  5  (a; -4)  =  4  (a; -3). 

60.  3  (a;  -  6)  -  2  (4  -  a;)  =  0. 

61.  7  (3  -  a;) -4  (7- 2a;)  =  0. 

62.  6(a;+ 5)- 12=:3(3a:- 1)  +4a;. 

63.  22  -  5  (3  —  2  a;)  =  a;  —  4  (a;  +  8). 

64.  8  (a;  -  7)  -  6  (a;  -  5)  =  5  (a;  -  4)  -  4  (a;  -  3). 

65.  (a;  +  1)^  =  a;^  +  5. 

66.  (a;-3)^  =  a;^-21. 

67.  (a;  +  Ay  =  a^(x  +  3). 

68.  (3a;+  l)^-2a;  =  9a;'+  13. 

69.  lx+  l)(a;+  2)  =  a;^  4-  11. 

70.  (x  +  3)(x  +  5)  =  a;^  +  31. 

71.  (a;  +  l)(a;  +  5)  =  (a;  +  2)(a;  +  3). 

72.  (x  -  10)(a;  -  7)  =  (a;  -  9) (a;  -  6). 

73.  (x  +  2Xx  +  4)  =  (a;  +  3)(a;  +  1)  +  1. 

74.  (x  -  4)(a;  +  1)  -  (a;  -  5)(a;  -  2)  =  0. 

75.  (x  -  6)(  X-  1)  -  (a;  +  7)(a;  +  3)  =  0. 

76.  (2a;+  l)(3a;+  l)  =  (6a;- l)(a;  +  2). 

77.  (16a;  -  5)(3a;  +  4)  =  (12a;  -  l)(4a;  +  3). 

78.  (2a;+  5)(5a;-4)  -  5a;  =  (10a;  -  3)(a;  +  1)  +  8. 

79.  ^x  =  S-^x.  89.  §a;  +  §a;  =  ^. 

80.  f  a;=:  1 -ia;.  90.  $a;  —  5  =  |  a;  —  4. 

81.  ^x  =  S+ix.  91.  h^  —  ^^+i^  =  h 

82.  ^a;  =  4  4-ia;.  92.  ^a;  +  ia;  +  ^a;  =  a;  ~  13. 

83.  I  a;  -  7  :=  i  a;.  93.  ^  (a;  +  3)  =  4. 

84.  ta;- 1  =  1 -^a;.  94.  ^  (a;  -  4)  =  1. 

85.  ^x  +  lx  =  2.  95.  I  (a;  +  1)  -  2  =  0. 

86.  ^x-^x  =  4:.  96.  i(a;  +  4)-ia;=:8. 

87.  ia;-ia;==12.  97.  ^  (x  +  l)  =  ^  (x  +  6). 

88.  h^  +  ^^  =  h  98.  |(5a;- l)-8  =  K4a^-2). 

99.  i(l  -  a;)  -  tV  (2  -  «^)  -=  tV  (3  +  a;). 

100.  i(a;  -  15)  -  T^ff  (9a;  -  2)  =  ia;  +  i. 

101.  ^(x  +  i)-i(x-'i)  =  10. 

102.  |(4a;-^)  +  ^(3a;-i)  =  TV. 

103.  ^(12a;-f)-f(14a;  +  ^)=:84. 

104.  la;  —  li  +  a;  =  ^  (6a;  -  9)  -  f,  a;. 


• 


174  FIRST  COURSE  IN  ALGEBRA 

!Equatioiis  in  which  Decimal  Fractions  appear  aniougf 
the  Coefficients 

Ex.  105.   .5  a;  =  .015. 

By  multiplying  both  members  of  the  equation  by  1000  we  shall  obtain  an 
equivalent  equation  in  which  the  decimal  coefficients  are  replaced  by 
integral  coefficients. 

That  is,  500  a:  =  15  Check.     .5  x  (.03)  =  .015 

Therefore,    x  =  ^^  .015  =  .016. 

Or,  X  =  .03,  in  decimal  notation. 

106.  .01  X  =  200.  113.  a;  —  .1  =  1  —  .liB. 

107.  .7  a;  =  .07.  114.  .2  a;  +  3  -  .04  a;  =  3.8. 

108.  .5a;=l.  115.  .2  a;  +  .04  =  .25  a;  -  .26. 

109.  .3  a;  =  3.  116.  .093  -  .1  a;  =  .02  a;  -  .13  a;  +  .01. 

110.  .2  a;  =  4.  117.  a;  —  10  +  .1  a;  =  1100  —  .01  a:. 

111.  .25a;=1.25.  118.  3 -. 2 a;  +  30  =  .02 a; -300  + .002a;. 


112.  .2  a;  =  48  — .04  a;. 


Problems 


7.  A  problem  is  a  question  proposed  for  solution. 

8.  A  problem  is  said  to  be  determinate  if  it  has  a  limited  or 
finite  number  of  solutions. 

In  the  contrary  case,  it  is  said  to  be  indeterminate. 

9.  To  solve  a  given  problem  is  to  find  the  values  of  certain 
unknown  quantities  whose  relations  with  one  another  and  with 
certain  known  quantities  are  given. 

The  relations  between  the  known  and  unknown  quantities  are 
called  the  conditions  of  the  problem. 

In  solving  a  problem  which  admits  of  algebraic  solution,  the  first 
step  to  be  taken  is  to  discover  the  relations  between  the  unknown 
and  the  known  quantities,  as  given  in  the  statement  of  the  problem. 

10.  The  beginner  will  find  it  helpful,  whenever  relations  are  given 
between  general  numbers,  to  consider  an  analogous  arithmetic 
problem,  choosing  definite  numerical  values  in  place  of  the  given 
general  numbers. 

E.  g.   By  how  much  does  a  exceed  15  ? 

Consider  a  similar  example  in  numbers.  By  how  much  does  21  ex- 
ceed 15? 


ALGEBRAIC  EXPRESSION  175 

The  excess  of  21  over  15  is  the  difference  between  21  and  15,  that  is, 
21  -  15. 

By  the  same  reasoning  it  appears  that  a  must  exceed  15  by  the  difference 
between  a  and  15,  that  is,  by  a  —  15. 

ALGEBRAIC   EXPRESSION^ 

Mental  Exercise  XL     2 

1 .  By  how  much  does  x  exceed  yl  • 

2.  By  how  much  does  a  exceed  6  ? 

3.  By  how  much  does  x  exceed  25  ? 

4.  By  how  much  does  30  exceed  y  ? 

5.  By  how  much  does  a  exceed  1  ? 

6.  What  number  must  be  added  to  x  to  obtain  a  ? 

7.  By  what  number  must  x  be  diminished  to  e<[ual  z  1 

8.  What  number  is  less  than  10  by  a  ? 

9.  What  number  is  greater  than  15  by  6  ? 

10.  If  a  represents  an  integer  how  may  the  next  greater  integer  be  repre- 
sented ?     The  next  less  ? 

11.  If  6  represents  an  odd  integer  how  may  the  next  greater  odd  integer 
be  represented  ?     The  next  less  ? 

12.  If  2  c  represents  an  even  integer  how  may  the  next  greater  even  in- 
teger be  represented  1 

13.  Find  an  expression  for  three  consecutive  integers  of  which  a  is  the 
least. 

14.  Find  an  expression  for  three  consecutive  integers  of  which  b  is  the 
greatest. 

15.  Find  an  expression  for  three  consecutive  integers  of  which  c  is  the 
one  between  the  other  two. 

16.  If  a  number  represented  by  x  is  separated  into  two  parts,  one  of 
which  is  5,  what  is  the  other  part  ? 

17.  If  a  number  represented  by  a  is  separated  into  two  parts  one  of 
which  is  6,  what  is  the  other  part  ? 

18.  Find  an  expression  for  the  greater  of  two  numbers  if  the  less  is  I  and 
the  difference  is  d. 

19.  A  man  sold  a  horse  for  ^h  and  gained  %  on  the  cost.  Find  an  ex- 
pression for  the  cost  of  the  horse. 

20.  A  boy  is  15  years  old  now.  Find  an  expression  for  his  age  x  years 
ago.     How  old  will  he  be  in  y  years  ? 

21.  If  a  boy  is  y  years  old  now  how  old  will  he  be  5  years  from  now  ? 
How  old  was  he  three  years  ago  ? 


176  FIRST  COURSP]   IN  ALGEBRA 

22.  A  man  has  $a  and  spends  $a:.     Find  an  expression  for  the  sum 
remaining. 

23.  In  uniform  motion,  Distance  =  Rate  x  Time. 

How  fai-  can  a  person  walk  in  a  hours  at  the  rate  of  b  miles  per  hour  ? 

24.  How  long  will  a  train  require  to  move  uniformly  a  distance  of  a 
miles  at  the  rate  of  x  miles  per  hour  ? 

25.  If  a  man's  expenses  are  §x  per  week  how  much  will  they  be  for  a 
year? 

26.  How  much  will  a  man  whose  wages  are  $a  per  day  earn  in  b  days  ? 
In  cd  days  ? 

27.  Express  $a  in  terms  of  cents. 

28.  Express  b  cents  in  terms  of  dollars. 

29.  Express  d  dimes  in  terms  of  dollars. 

30.  Express  ^  and  b  dimes  in  terms  of  cents. 

31.  Express  x  half  dollars  and  y  quarters  in  terms  of  cents. 

32.  Express  y  yards  in  terms  of  feet. 

33.  Express /feet  in  terms  of  inches. 

34.  Express  h  inches  in  terms  of  yards. 

35.  Express  m  inches  in  terms  of  feet. 

36.  Express  x  feet  plus  y  inches  in  terms  of  inches. 

37.  Express  a  yards  plus  b  feet  plus  c  inches  in  terms  of  inches. 

38.  Find  an  expressiou  for  a  gallons  in  terms  of  quarts.     In  terms  of 
pints. 

39.  Find  an  expression  for  x  pounds  in  terms  of  ounces  and  of  t  tons  in 
terms  of  pounds. 

40.  Find  an  expression  for  a  gallons  plus  &  quarts  in  terms  of  quarts. 

41.  Find  an  expression  for  x  pounds  plus  y  ounces  in  terms  of  ounces. 

42.  Find  an  expression  for  a  acres  in  terms  of  square  rods. 

43.  Find  an  expression  for  x  acres  plus  y  square  rods  in  terms  of  square 
rods. 

44.  Express  h  hours  in  terms  of  minutes. 

45.  Express  a  minutes  in  terms  of  hours.     In  terms  of  seconds. 

46.  Express  a  days  plus  b  hours  in  terms  of  hours. 

47.  Express  m  hours  plus  n  minutes  in  terms  of  seconds. 

48.  What  is  the  cost  of  m  books  at  n  cents  each  ? 

49.  If  b  books  cost  ^  find  an  expression  for  the  cost  of  one  book. 

50.  If  one  book  cost  c  cents  how  many  cents  do  b  books  cost  ?     How 
many  dollars  ? 

51.  If  the  interest  on  $1  for  one  year  is  r  cents  what  will  be  the  interest 
on  ^  for  one  year  at  the  same  rate  ] 

52.  Find  an  expression  for  one  per  cent  of  a  ;  five  per  cent  of  b  ;  seventy 
five  per  cent  of  c  ;  thirty-seven  and  one-half  per  cent  of  d. 


PROBLEMS  177 

53.  Find  an  expression  for  a  per  cent  of  h. 

54.  Find  an  expression  for  x  per  cent  of  x. 

55.  Find  an  expression  in  square  feet  for  the  area  of  a  rectangular  room 
which  is  a  feet  long  and  b  feet  wide.     In  square  yards. 

56.  What  is  the  area  in  square  feet  of  a  square  room  each  of  whose  sides 
is  X  feet  in  length  ?     In  square  3'ards  1 

57.  Find  an  expression  in  square  yards  for  the  area  of  a  rectangular 
room  which  is  x  yards  long  and  y  yards  wide.     In  square  feet. 

Exercise  XL    3 

Translate  the  following  statements  into  algebraic  language  and 
express  the  given  conditions  by  means  of  conditional  equations,  and 
solve.  The  solutions  of  the  equations  should  in  every  case  be  ex- 
amined to  see  if  they  satisfy  the  conditions  of  the  given  problems  : 

1.  Find  two  numbers  whose  sum  is  46,  which  are  such  that  the  greater 
number  shall  exceed  the  less  by  12. 

In  this  problem  the  values  of  the  two  numbers  required  are  unknown, 
but  by  the  statement  we  know  that  if  ar  represents  the  less  number,  a;  +  12 
must  represent  the  greater. 

Since  the  sum  of  the  numbers  is  46,  we  have 

(first  number)  +  (second  number)  =  46. 
Hence,  x  +  (a:  +  12)         =46. 

Solving,  we  find  that  x  =  17. 

Accordingly  the  less  number  is  17  and  the  greater  number  which  is 
represented  by  a:  -f-  12  must  be  29. 

These  numbers  satisfy  the  conditions  of  the  problem. 

2.  Find  a  number  which,  when  multiplied  by  7,  exceeds  36  by  as  much 
as  the  number  itself  falls  short  of  36.  , 

Let  X  represent  the  required  number. 

Translating  the  conditions  of  the  given  problem  into  algebraic  language, 
we  obtain  the  conditional  equation 

7  a:  -  36  =  36  -  a:, 
whose  solution  is  found  to  be  a:  =  9,  which  is  the  required  number. 

It  will  be  found  that  9  satisfies  the  conditions  of  the  given  problem. 

Suggestions  for  stating  and  solving  simple  problems. 

Fi?'st,  decide  what  unknown  number  is  to  be  found,  and  represent 
it  by  some  letter,  say  ^. 

It  should  be  observed  that  the  letters  appearing  in  conditional 
equations  can  represent  abstract  numbers  only, 

12 


178  FIRST  COURSE  IN  ALGEBRA 

E.  g.  The  letter  z  cannot  represent  a  given  sum  of  money  or  a  given 
distance,  but  it  may  stand  for  the  number  of  dollars  in  a  given  sum,  or  the 
number  of  units  of  length,  say  feet,  yards,  or  miles,  in  terms  of  which  a 
given  distance  is  expressed. 

It  is  essential  that  we  should  be  consistent  in  expressing  the 
known  and  unknown  numbers  of  the  problem  in  terms  of  the  same 
unit. 

E.  g.  If  the  unknown  letter  represents  the  number  of  miles  in  a  given 
distance,  then  the  remaining  numbers  in  the  conditional  equation  must  be 
expressed  in  terms  of  miles. 

Second,  examine  the  statement  of  the  problem  to  discover  two 
independent  conditions  which  may  lead  to  two  different  algebraic 
expressions  for  the  same  number. 

Third,  obtain  Uvo  algebraic  expressions  for  this  number. 

Fourth,  use  these  expressions  as  members  of  a  conditional  equa- 
tion, and  solve. 

Fifth,  examine  the  solutions  of  the  conditional  equation,  and 
determine  whether  or  not  they  satisfy  the  stated  conditions  of  the 
problem. 

3.  Find  two  numbers  whose  sum  is  28  and  whose  difference  is  6. 

4.  Find  three  consecutive  numbers  whose  sum  is  42. 

Suggestion.  Let  x  represent  the  first  number,  a:  +  1  the  second  number, 
and  X  +  2  the  third  number. 

5.  Find  three  consecutive  numbers  whose  sum  is  96. 

6.  Find  four  consecutive  numbers  whose  sum  is  206. 

7.  Find  three  consecutive  even  numbers  whose  sum  is  54. 

8.  Find  two  consecutive  numbers  such  that  seven  times  the  first  shall 
exceed  four  times  the  second  by  29. 

9.  Find  a  set  of  six  consecutive  numbers  such  that  the  sum  of  the  first 
and  last  shall  be  95. 

10.  What  number  is  it  whose  double  is  25  more  than  its  third  part  ? 

11.  Find  two  numbers  differing  by  12  whose  sum  is  twice  their 
difference. 

Let  X  represent  the  greater  number. 
Then  a;  —  12  represents  the  less  number. 
We  have  a;  +  (x  -  12)  =  2  •  12 

From  which  a:  =  18. 


PROBLEMS  179 

Accordingly  the  greater  number  is  18,  and  the  less  number,  represented 
by  a;  -  12,  is  6. 

These  numbers  will  be  found  to  satisfy  the  conditions  of  the  given 
problem. 

12.  Find  two  numbers  whose  sum  is  36  and  whose  difference  is  10. 

13.  Separate  63  into  two  parts  such  that  one  part  shall  be  greater  than 
the  remaining  part  by  5. 

14.  Separate  147  into  two  parts  such  that  the  greater  part  shall  exceed 
tlie  less  by  7  more  than  ^  of  the  less  part. 

15.  Separate  72  into  two  parts  such  that  three-fourths  of  the  less  part 
shall  exceed  three-eighths  of  the  greater  by  9. 

16.  Find  a  number  such  that  if  one  be  added  to  three-fourths  of  the 
number  the  result  will  be  four  more  than  one-half  of  the  number. 

17.  One-half  of  a  certain  number  exceeds  the  sum  of  its  fifth  and  sixth 
parts  by  4.     Find  the  number. 

18.  The  sum  of  the  third  and  fourth  parts  of  a  certain  number  exceeds 
the  sum  of  the  fifth  and  sixth  parts  by  13.     Find  the  number. 

19.  If  5  X  -f-  4  stands  for  49,  for  what  number  will  x  —  2  stand  ? 

On  condition  that  5  x  +  4  shall  represent  49,  it  is  necessary  that  x  shall 
satisfy  the  conditional  equation  5  a:  -|-  4  =  49. 

The  solution  of  this  equation  is  found  to  be  a:  =  9. 
Accordingly,  the  expression  x  —  2  must  represent  7. 

20.  The  sum  of  the  two  digits  of  a  number  is  11.  If  27  be  added  to  the 
number  the  digits  will  be  interchanged.     What  is  the  number  ? 

Let  X  represent  the  digit  in  units'  place. 

It  follows  that  11  —  a;  represents  the  digit  in  tens'  place. 

The  given  number  may  be  represented  by 

10  (digit  in  tens'  place)  -\-  (digit  in  units'  j)hvce), 
that  is,  by  10  (11  —  a;)  -f  x. 

If  the  digits  are  interchanged,  the  resulting  number  will  be  represented  by 
10a:-f(ll -a:). 

By  the  conditions  of  the  problem,  we  have 

10  (11  -  ar)  +  a:  -f  27  =  10  a:  -f  (11  -  a:). 

Solving,  we  find  that  x  =  7. 

Hence  the  digit  in  units'  place  is  7,  and  the  digit  in  tens'  place,  which 
is  represented  by  11  —  a:,  must  be  4. 

Accordingly,  the  required  number  is  4  x  10  -|-  7,  that  is,  47. 

It  may  be  seen  that  47  satisfies  the  conditions  of  the  problem,  for  the 
sum  of  the  digits  is  11,  and  if  27  be  added  to  47  the  resulting  number  is  74, 
which  may  also  be  obtained  by  interchanging  the  digits  of  47. 


180  FIRST  COURSE  IN  ALGEBRA 

21.  The  sum  of  the  two  digits  of  a  number  is  12.  If  36  be  added  to 
the  number  the  digits  will  be  interchanged.     What  is  the  number '? 

22.  The  sum  of  the  three  digits  of  a  number  is  9,  and  the  digit  in  hun- 
dreds' place  is  three  times  that  in  units'  place.  If  180  be  added  to  the 
number,  the  digits  in  hundreds'  and  tens'  places  will  be  interchanged. 
What  is  the  number? 

23.  Find  such  a  number  that,  when  used  to  diminish  each  of  the  two 
indicated  fiictors  of  the  two  unequal  products  45  •  75  and  51-66,  the  result- 
ing products  will  be  equal. 

24.  A's  age  exceeds  B's  by  14  years.  Eight  years  ago  A  was  three 
times  as  old  as  B.     Find  the  present  age  of  each. 

Let  b  represent  the  number  of  years  in  B's  present  age. 

Then,  by  the  conditions  of  the  problem,  the  number  of  years  in  A's 
present  age  is  represented  by  6  -}-  14. 

The  numbers  of  years  in  A's  and  B's  ages  eight  years  ago  would  accord- 
ingly be  represented  by  (6  +  14)  —  8  and  6  —  8,  respectively. 

Since  at  that  time  A's  age  was  three  times  that  of  B's,  we  may  form  the 
conditional  equation 

(6-f  14)-'8  =  3(6-8). 

Solving,  we  obtain  6  =  15. 

AccoKlingly,  B's  present  age  is  15  years,  and  A's  present  age,  which  is 
represented  by  6  -j-  14,  must  be  29  years. 

These  numbers  are  found  to  satisfy  both  the  statement  of  the  problem 
and  the  algebraic  equation. 

25.  The  ages  of  A  and  B  are  such  that  five  years  ago  A's  age  was  four 
times  that  of  B's,  while  five  years  hence  his  age  will  be  twice  that  of  B's. 
Find  the  present  age  of  each. 

26.  The  sum  of  the  ages  of  A  and  B  is  36  years,  and  six  years  hence 
A's  age  will  be  three  times  that  of  B's.     Find  their  present  ages. 

27.  In  a  company  of  22  persons  a  resolution  is  carried  by  a  majority  of 
12,  all  voting.     How  many  voted  for  the  measure  ? 

28.  In  an  informal  ballot  a  resolution  was  adopted  by  a  majority  of  six 
votes,  but  in  a  formal  vote  one-third  of  those  who  had  before  voted  for  it 
voted  against  it,  and  the  resolution  was  lost  by  a  majority  of  four  votes. 
How  many  voted  each  way  in  the  formal  ballot? 

29.  A  grocer  estimated  that  his  supply  of  sugar  would  last  eight  weeks. 
He  sold  on  an  average  50  pounds  a  day  more  than  he  expected.  It  lasted 
him  six  weeks.     How  much  did  he  have  ? 

30.  Determine  how  an  amount  of  $135  must  be  divided  among  three 
persons  in  such  a  way  that  the  share  of  the  first  shall  be  three  times  that  of 
the  second,  and  the  share  of  the  second  twice  that  of  the  third. 


PROBLEMS  181 

Let  X  represent  the  number  of  dollars  in  the  share  of  the  third.  Then  the 
number  of  dollars  in  the  shares  of  the  second  and  tirst  will  be  represented 
by  2  X  and  6  x  respectively. 

By  the  conditions  of  the  problem,  we  may  construct  the  conditional  equa- 
tion a;  +  2a:  +  6a:=  135,  whose  solution  is  found  to  be  a;  =  15. 

Accordingly,  the  share  of  the  third  is  $15 ;  the  share  of  the  second  repre- 
sented by  2  a:,  is  $30 ;  and  the  share  of  the  first,  represented  by  6  a;,  is  $90. 

These  amounts  are  found  to  satisfy  the  conditions  of  the  given  problem. 

31.  A  sum  of  $7924  was  bequeathed  to  three  persons  with  the  stipulation 
that  the  first  was  to  receive  twice  as  nuich  as  the  second  and  one-half  as 
much  as  the  third.     Determine  the  amounts. 

32.  A  man  wishes  to  divide  the  sum  of  $99  into  five  parts  in  such  a  way 
that  the  first  part  shall  exceed  the  second  by  $3,  be  less  than  the  third  part 
by  $10,  greater  than  the  fourth  part  by  $9  and  less  than  the  fifth  part  by 
$16.     Find  the  parts. 

33.  A  paymaster,  wishing  to  use  $25,662  on  pay-day,  requested  the  pay- 
ing teller  to  make  up  the  amount  in  tlie  following  way:  A  certain  number 
of  $100  bills,  three  times  as  many  fifties,  four  times  as  many  twenties  as 
fifties,  twice  as  many  tens  tis  fifties,  three  times  as  many  fives  as  tens,  as  many 
twos  as  tens,  as  many  ones  as  twos. 

How  many  bills  of  each  denomination  were  given  ? 

34.  At  two  stations,  A  and  B,  on  a  line  of  railway,  the  prices  of  coal  are 
$3.50  per  ton  and  $4  per  ton  respectively.  If  the  distance  between  A  and  B 
be  150  miles  and  coal  can  be  shipped  for  one-half  a  cent  per  ton  per  mile, 
find  the  place  on  the  railway  between  A  and  B  at  which  it  will  be  indiffer- 
ent to  a  customer  whether  he  buys  coal  from  A  or  from  B. 

35.  A  farmer  estimated  that  his  supply  of  feed  for  Ids  50  cows  would 
last  only  12  weeks.  How  many  cows  must  he  sell  in  order  that  the  supply 
may  last  20  weeks  ? 

36.  A  contractor  undertakes  to  complete  a  certain  amount  of  work  in  a 
given  time.  By  the  terms  of  the  contract  he  is  to  receive  $12  for  each  day's 
work  during  the  given  time,  and  is  to  forfeit  $5  for  each  day  taken  beyond 
that  time.  If  the  total  amount  received  was  $167  for  21  days'  work,  find 
the  time  for  the  original  contract. 

37.  It  was  estimated  that  a  certain  amount  of  earth  could  be  excavated 
by  a  steam  shovel  alone  in  12  days,  or  by  a  gang  of  laborers  alone  in  28 
<lay8.  After  being  used  a  certain  number  of  days  the  shovel  was  disabled, 
and  the  work  was  then  completed  by  the  men,  who  worked  2  days  less  than 
the  time  during  which  the  shovel  had  been  used.  During  how  many  days 
was  the  shovel  used  ? 

38.  Sixty  laborers  were  engaged  to  remove  an  embankment.     Some  of 


182  FIRST  COURSE  IN  ALGEBRA 

them  were  engaged  at  the  rate  of^l.lO  a  day,  and  the  others  at  the  rate 
of  $1.60  a  day.  The  memorandum  having  been  lost,  it  was  rec^uired  to  find 
how  many  worked  at  each  rate,  if  the  total  amount  paid  was  ^0. 

39.  The  help  of  a  certain  factory,  numbering  31G,  consists  of  men  and 
boys.  If  the  weekly  pay  of  each  man  is  ^12  and  that  of  each  boy  ^i  thid 
the  number  of  each,  if  the  weekly  pay  roll  amounts  to  $2688. 

40.  It  is  observed  that  a  square  room  requires  one  and  one-ninth  square 
yards  less  of  carpeting  than  a  rectangular  room  whose  length  is  one  yard 
longer,  and  width  two  feet  less,  than  the  side  of  the  square  room.  Find 
the  area  of  each  of  the  rooms. 

41.  A  man  has  $4200  in  four  banks.  He  has  twice  as  much  in  the 
second  bank  as  in  the  first,  as  nmch  in  the  third  as  in  the  first  and  second 
together,  and  twice  as  much  in  the  fourth  as  in  the  first  and  third  together. 
How  nmcli  money  has  he  in  the  fourth  bank  ? 

42.  A  man  invests  \  of  liis  capital  at  4  per  cent,  \  at  3  per  cent,  and  the 
remainder  at  3^  per  cent,  and  thus  secures  an  annual  income  of  $1390. 
What  is  his  capital  ? 

43.  A  man  invests  ^  of  his  capital  at  3  per  cent  and  the  rest  at  3^  per 
cent,  and  thus  receives  an  annual  income  of  $55.50.     What  is  his  capital  1 

44.  A  man  desires  to  invest  his  capital  of  $10,000,  partly  at  5  per  cent 
and  partly  at  3  per  cent.  How  must  he  invest  the  amounts  so  that  his 
yearly  income  shall  be  at  the  rate  of  4j  per  cent  ? 

45.  How  may  $3600  be  invested,  partly  in  5  per  cent  stock  whose  market 
value  is  eighty  cents  on  a  dollar,  and  the  remainder  in  6  per  cent  stock 
selling  for  one  dollar  and  twenty  cents  on  a  dollar,  in  order  that  the  income 
from  the  two  sources  may  be  the  same  ? 

46.  How  long  will  it  take  an  investment  of  $5730  to  amount  to  $6589.50 
at  3  per  cent  simple  interest  ? 

47.  At  what  time  between  three  and  four  o'clock  are  the  hands  of  a 
watch  together  ? 

Let  X  represent  the  number  of  minute  spaces  passed  over  by  the  minute- 
hand,  reckoned  from  the  position  occupied  by  the  hand  at  3  o'clock,  to  the 
point  where  it  overtakes  the  hour-hand.  Then,  during  the  same  time,  the 
hour-hand  will  pass  over  a:/12  minute  spaces. 

Since  at  3  o'clock  the  hour-hand  is  15  minute  spaces  ahead  of  the  minute- 
hand,  it  follows  that  the  distance  x  passed  over  by  the  minute-hand  must 
be  greater  by  15  minute  spaces  than  the  distance  a;/12  passed  over  by  the 
hour-hand. 

Accordingly,  we  have  x  —  a:/12  =  15, 

from  which  we  obtain  x  =  16y\,  which  is  the  number  of  minutes  after  3 
o'clock  when  the  hands  of  the  watch  are  together. 


PROBLEMS  18  B 

48.  At  what  time  between  9  and  10  o'clock  are  the  hands  of  a  watch 
opposite  each  other  ? 

49.  At  what  time  between  7  and  8  o'clock  will  the  hands  of  a  clock 
be  at  right  angles  to  each  other  ? 

50.  At  what  time  between  5  and  6  o'clock  is  the  minute  hand  of  a 
watch  two  minutes  in  advance  of  the  hour  hand  1 

51.  Two  trains  start  at  the  same  time  from  different  stations  400  miles 
apart.  One  travels  at  the  rate  of  48  miles  an  hour  and  the  other  at  the 
rate  of  32  miles  an  hour.  How  far  does  the  faster  train  travel  before  meet- 
ing the  slower  one  1 

52.  Two  hours  after  a  train  left  a  certain  station  a  second  train  was 
dispatched,  and  it  overtook  the  first  train  in  four  hours.  To  accomplish  this 
it  was  necessary  for  the  second  train  to  run  15  miles  an  hour  faster  than 
the  first.     How  many  miles  per  hour  did  the  trains  run  ? 

53.  Where  must  a  side  track  be  placed  on  a  single-track  railway  in  order 
that  an  express  train,  travelling  at  the  rate  of  46  miles  an  hour,  may  not  be 
delayed  by  an  accommodation  train  travelling  toward  it  at  the  rate  of  29  miles 
an  hour,  the  two  trains  stai'ting  at  the  same  time  from  two  places  60  miles 
apart  ? 

54.  If  an  outward  trip  of  an  excursion  train  is  made  at  the  rate  of  20 
miles  an  hour  and  the  return  trip  at  16  miles  an  hour,  the  whole  time  being 
9  hours,  what  is  the  distance  ( 

55.  Two  ships  start  from  a  given  port  at  the  same  time,  one  going  north 
at  the  rate  of  11  miles  per  hour  and  the  other  going  south  at  the  rate  of  7 
miles  per  hour.  How  long  after  starting  will  they  at  these  rates  be  exactly 
108  miles  apart  ? 

56.  Two  ferry  boats,  whose  rates  are  15  miles  and  12  miles  an  hour 
respectivel}',  start  simultaneously  from  opposite  shores  of  a  river,  three- 
fourths  of  a  mile  wide.     Where  will  they  meet  ? 

Problems  in  Science 

57.  A  certain  grade  of  sulphuric  acid  is  known  to  be  94  per  cent  pure. 
How  much  distilled  water  must  be  added  to  a  gallon  of  this  sulphuric  acid 
in  order  that  the  mixture  may  be  75  per  cent  pure  1 

Represent  by  x  the  number  of  gallons  of  distilled  water  to  be  added. 
Then  the  number  of  gallons  of  the  mixture  will  be  represented  by  1  -j-  x. 

Since  75  per  cent  of  this  diluted  solution  is  pure  acid,  we  have  as  an 
expression  for  the  number  of  gallons  of  acid  .75  (I  +  x). 

By  adding  water,  the  amount  of  acid,  which  was  given  as  94  per  cent  of 
the  original  ''allon,  has  been  neither  increased  nor  decreased. 


184  FIRST  COURSE  IN  ALGEBRA 

Hence  we  may  use  these  two  expressions  for  the  same  amount  of  acid  as 
the  members  of  a  conditional  equation,  as  follows  : 

.75(1  +a;)  =  .94, 
from  which  we  obtain  x  —  .253 J. 

Accordingly  it  is  necessary  to  add  .253^  gallons  of  distilled  water. 

It  may  be  seen  that  this  amount  of  water  satisfies  the  condition  of  the 
problem  as  stated,  for  by  the  addition  of  .253^  gallons  of  wsiter  the  mixture 
is  made  to  consist  of  1.253|  gallons,  and  75  per  cent  of  this  mixture,  that  is 
94  per  cent  of  one  gallon,  is  pure  acid. 

58.  How  much  water  must  be  added  to  a  pint  of  alcohol  85  per  cent 
pure,  in  OKler  to  make  the  mixture  three-fourths  water  ? 

69.  How  much  12  per  cent  solution  of  a  certain  chemical  must  be  added 
to  a  gallon  of  4  per  cent  solution  to  raise  it  to  a  6  per  cent  solution  ? 

60.  How  much  water  must  be  added  to  6  quarts  of  acid  which  is  10  per 
cent  of  full  strength  to  make  the  mixture  8^  per  cent  of  full  strength  ? 

61.  How  much  water  must  be  added  to  a  gallon  of  three  per  cent  solu- 
tion of  a  certain  chemical  to  reduce  it  to  a  one  per  cent  solution  ? 

62.  How  many  ounces  of  pure  silver  must  be  melted  with  300  ounces  of 
silver  600  fine  in  order  to  make  a  bar  of  metal  800  fine  ? 

By  600  fine  is  meant  the  number  of  parts  \\\  1000  which  are  pure  metal. 

63.  How  many  ounces  each  of  two  bars  of  silver  which  are  800  fine  and 
725  fine  respectively  must  be  melted  together  to  make  a  bar  of  GO  ounces 
which  shall  be  775  fine  ? 

64.  How  many  pounds  of  pure  copper  must  be  melted  with  500  pounds  of 
gold  67/72  pure  in  order  to  make  the  composition  9/10  pure  gold  ? 

65.  It  has  been  said  that  the  crown  of  Hiero  of  Syracuse,  which  was 
part  gold  and  part  silver,  weighed  20  pounds  in  the  air  and  18|  pounds 
when  weighed  while  immersed  in  water.  Find  how  much  gold  and  how 
much  silver  it  contained,  knowing  that  19^  pounds  of  gold  and  10^  pounds 
of  silver  each  lose  one  pound  when  weighed  immersed  in  water. 

66.  The  pendulum  of  a  clock  swings  364  times  in  5  minutes  while  that 
of  a  second  clock  swings  233  times  in  4  minutes.  After  how  long  will  the 
second  have  swung  582  times  more  than  the  first  ? 


INTEGRAL  FACTORS  185 


CHAPTER  XII 
FACTORS  OF  RATIONAL  INTEGRAL  EXPRESSIONS 

1.  From  the  point  of  view  of  multiplication,  the  numbers  or 
expressions  which  are  multiplied  together  to  form  a  product  are 
called  factors  of  the  product. 

E.  g.    Since  2  x  5  =  10,  2  and  5  are  factors  of  10. 

Since  2  (m  +  ?i)  =  2  m  +  2  w,  2  and  w  +  w  are  factors  of  2  ?7i  +  2  w. 

Similarly  a  .+  6  and  a  —  b  are  factors  of  a^  —  b^. 

Also,  X  -\-  5  and  x  +  5  are  factors  of  x^  +  lOx  ■\-  25. 

2.  Our  present  problem  of  factoring  is  that  of  finding  two  or 
more  numbers  or  expressions  which  may  be  multiplied  together  to 
produce  a  given  expression  as  a  product. 

3.  Since  for  the  present  it  will  seldom  be  necessary  to  use  factors 
of  given  expressions,  except  such  as  are  altogether  integral  and 
rational,  we  shall  in  this  chapter  consider  only  integral  rational 
expressions,  and  shall  restrict  the  word  "  factor  "  to  mean  only  such 
a  factor  as  is  entirely  integral  and  rational. 

4.  An  integral  rational  algebraic  expression  is  said  to  be  prime 
if  it  has  no  exact  integral  rational  divisors  except  itself  and  unity 
(positive  or  negative). 

E.  g.    The  following  expressions  are  algebraically  prime  : 

a,  6  +  C,  X2+7/2+  1. 

5.  An  integral  rational  algebraic  expression  which  is  not  prime 
is  said  to  be  composite. 

E.  g.    The  following  expressions  are  composite  : 
a^,  6^  —  c^,  rnx^  4-  my*  +  m. 

6.  When  completely  factored,  all  of  the  prime,  integral,  rational 
divisors  of  a  given  polynomial  expression  are  made  to  appear,  and 
a  chain  of  additions  and  subtractions  is  transformed  into  one  of 
multiplications. 


186  FIRST  COURSE  IN  ALGEBRA 

E.  g.  The  number  represented  by  the  chain  of  atUlitions  10  a^  -f  70  a  -}- 
120  may   be  expressed   also   by  the  following  chain  of  multiplications: 

2  •  5  (a  +  3)(a  +  4).  Check,   a  =  2. 

That  is,         10  a^  +  70 a  4-  120  =  2 . 6  (a  +  3)(a  +  4).  300  =  300. 

Hence  it  appears  that  the  prime,  integral,  rational  factors  or  divisors  of 
10  a2  +  70  a  +  120  are  2,  5,  a  +  3,  and  a  +  4. 

7.  The  separate  factors  of  an  expression  are  never  of  degree 
higher  than  that  of  the  given  expression. 

8.  The  whole  number  of  algebraic  expressions  which  may  be 
written  is  unlimited,  and  of  these  some  do  not  admit  of  being 
expressed  as  the  product  of  two  or  more  integral  rational  factors. 

Hence,  in  many  cases  it  is  impossible  to  tell  beforehand  whether 
or  not  integral  rational  factors  of  a  given  expression  exist,  or  to 
determine  their  degrees  even  if  they  do. 

9.  Since  factoring  is  commonly  used  as  a  means  for  abbreviating 
algebraic  work,  it  is  necessary  for  the  beginner  to  become  thoroughly 
familiar  with  certain  methods  which  may  be  employed  in  elementary 
work. 

10.  In  examining  an  expression  for  factors,  it  is  natural  to  deter- 
mine first  what  factors,  if  any,  are  common  to  all  of  the  terms. 

Expressions  in  which  all  of  the  Terms  contain  a  Common 
Monomial  Factor 

11.  1/  some  letter  is  common  to  all  0/  the  terms  of  an  expression 
it  follows  that  every  term^  and  accordingly  the  whole  expression,  is 
exactly  divisible  by  that  letter. 

This  follows  from  the  converse  of  the  distributive  law  of 
multiplication, 

oca  ■¥  xh  +  xc  + =  x{a  +  6  +  c  + ). 

12.  All  expressions  should  be  examined  first  for  common  factors, 
for  it  often  happens,  after  removing  a  common  factor,  that  the  form 
of  the  resulting  expression  is  such  that  some  other  process  may  be 
then  applied  to  simplify  further  the  factors  which  may  thus  be 
obtained. 

Ex.  1.   Factor  20  a^  +  15  ««  +  10  a. 

Observe  that  the  monomial  factor  ba  is  found  in  every  term,  and  that 
this  is  all  that  is  common  to  all  of  the  terms. 


INTEGRAL  FACTORS  187 

Using  5  a  as  a  divisor,  th&  quotient  obtained  is  4  a^  +  3  a  +  2. 
Hence  20 a^  +  15 a^  +  10 a  =  5 a{4a^  +  3 a  +  2).     Check.  Let  a  =  2. 

160  +  60  -f.  20  =  10  (16  +  6  +  2) 
240  =  240. 
It  appears  that  the  factors  of  the  given  expression  are  5,  a,  and  the  ex- 
pression 4  a^  +  3  a  -f  2. 

Ex.  2.    Factor  44  a%^  -  22  a%\ 

Since  22a^6^  is  all  that  is  common  to  both  terms,  we  have 

Check.     Let  a  =  2,  6  =  1. 
44a%^  -  22a36*  =  22a%\2a  -h).  44  •  16  -  22  •  8  =  22  •  8  (4  -  1) 

528  =  528. 
The  different  prime  factors  of  the  given  expression  are  2, 11,  a,  and  6,  and 
the  binomial  (2  a  —  h). 

Exercise  XII.     1 

Obtain  factors  of  the  following  expressions,  verifying  all  results 
by  numerical  checks  : 

1.  ah  +  ac.  10.  a^  +  a.-^  +  x\ 

2.  bed  +  bee.  11.  5  a^b''  +  3  ^^^  -  4  a''b\ 

3.  3ic  +  3y+32;.  12.  Ix'-Ua?  +  2\x\ 

4.  ^xij  ^-  \2zy  +  ZOwy.  13.  9a^c  +  ISac^  —  9ac. 

5.  14ic  +  14.  14.  ^a'^bc  +  ^ab'^c  +  3abc\ 

6.  4  a;»  +  4  if''.  15.  mhi"  -  3  m^/i^*  -  mhi\ 

7.  5  a'  -  6  a6.  16.  a^b''  -  b\  ~  b^'d^  +  i^. 

8.  9a;^  -  3x.  17.  5a^c^  +  lOac^e  -  5ar/. 

9.  8  a'^^  -  8  ab\  18.  14  a^/^'^  +  28  a^b""  -  7  a2<^«. 

19.  20*^=^V  +  25rtm*=  -  3()rt'(^'c^ 

20.  38  xY  +  57  a-Y  -  19  x\ 

21.  34  a^b^  -  51  a«/^o  +  85  a^bc\ 

22.  32  icY;?''  -  48  xfz^  +  64  xYz. 

23.  77  w^;^r  —  99  m^n  +  88  wrs^ 

24.  13  a^gx^  +15  tt^V  —  2  aVic  —  9  agx. 

25.  a^^c^c^  +  a^/^c^^^  +  a'b'cd'  +  a^^Vc^. 

26.  xYz^  +  xYz^  +  xY^""' 

27.  a%''cH^ -^  ab^'c^'d^  +  ah''c'd\ 

28.  3  icy;^  -  6  xifz''  +  3  xY""- 

29.  ffa!'"3r  -  ^a;"^/'"  +  C£c'"y'"  -  ^/a-"*?/"*. 

30.  a'"+^6'"+^c'"  —  a'"6"*+^c'"+^  +  a"'+^6"*c"^+\ 


188  FIRST  COURSE  IN  ALGEBRA 

13.  If  all  of  the  terms  of  a  given  expression  do  not  contain  a 
common  factor,  it  may  be  possible  to  so  arrange  them  in  groups 
that  there  shall  be 

Groups  of  Terms  Having  a  Common  Factor 

The  process  for  factoring  such  expressions  may  be  obtained  by 
applying  the  converse  of  the  law  of  distribution  for  multiplication. 
Consider  the  identity 

(a  +  b){x  ■\-y-{-z)  =  {a  +  b)x+{a-\-  h)y  +  (a  +  b)z 
=  ax  +  bx-]-ai/  +  bi/  +  az-\-  bz. 

By  reversing  the  steps  in  this  process  the  last  expression,  which 
consists  of  a  chain  of  additions,  may  be  transformed  into  the  origi- 
nal expression,  which  is  the  product  of  two  factors. 

14.  The  complete  distributed  product  of  the  sum  of  two  terms,  a  and  6, 
multiplied  by  the  sum  of  three  other  terms,  x,  y,  and  z,  different  from  the 
first,  will  contain  2  x  3  or  6  terms,  ax,  bx^  ay,  by,  az  and  bz. 

Consequently,  when  factoring  ax  -{■  bx  +  ay  -}- by  -{- az  -\-  bz,  we  may  re- 
verse the  process,. and  when  seeking  groups  of  terms  which  have  common 
factors,  we  may  look  for  either  three  groups  of  two  terms  each  or  two 
groups  of  three  terms  each. 

Ex.  1.  Factor  (a^  +  2)(m  -  n)  -  (3  a^  -  6)(m  -  n). 

Since  the  factor  m  —  n  occurs  in  both  terms  we  may  add  with  respect 
to  m  —  n  as  a  summand,  and  obtain 

(a2  -f  2)(m  -  7i)  -  (3  a2  -  6)(m  -  n)  =  (a^  +  2  -  3 a^  +  6)(m  -  n) 

=  (2  -2a^  +  b)(m-n). 
Check.     Let  a  =  2,  b  =  2,  m  =  5,  and  n  =  4. 
(4  +  2)  (5  -  4)  -  (12  _  3)  (5  -  4)  =  (2  -  8  +  3)  (5  -  4) 

-  3  =  -  3. 
Ex.  2.    Factor  ax  -  bay  i- 2bx  -  10 by. 

There  is  no  monomial  factor  common  to  all  of  the  terms,  but  there  are 
groups  of  terms  which  have  common  monomial  factors. 

Since  the  expression  contains  four  terms,  we  may  expect  to  find  that  it 
can  be  separated  into  a  product  of  two  factors,  each  of  which  contains  two 
terms. 

The  factor  a  is  found  to  be  common  to  the  two  terms  ax  and  —  5  ay, 
and  the  factor  2  6  is  common  to  the  two  terms  2  bx  and  —  10  by. 


INTEGRAL  FACTORS  189 

Separating  the  monomial  factors,  a  and  2  b,  by  division,  from  the  terms 
containing  them,  we  obtain 

ax  —  5  ay  +  2bx  —  lOhj  —  a(x  —  5y)  +  2h(x  —  b  y). 

The  binomial  factor  x  —  by  is  common  to  the  terms  a{x  —  by)  and 
2b{x-by). 

Hence,  using  a;  —  5  ?/  as  a  divisor,  the  expression  a(x  —  by) -{-^bi^x  —  by) 
may  be  written  in  the  form  (a  +  2  b){x  —  by). 
The  work  may  be  arranged  as  follows  : 
ax  —  bay  -\-2bx  —  10 by  =  a{x  —  bij)  +  2b{x  —  by)  Check. 

Let  a  =  2,  6  =  3,  a;  =  6,  2/  =  1. 
=  (a  +  2b)(x  -by).      12  -  10  +  36  -  30  =  8  •  1 

8  =  8. 

The  same  result  can  be  obtained  by  grouping  the  terms  ax  and  2  bx,  and 
also  —  bay  and  —  10 by. 

The  student  should  carry  out  the  work  with  this  grouping. 

15.  It  should  be  remembered  that  an  e.rpression  is  not  factored, 
unless  it  is  ivritten  as  a  single  product,  not  as  the  sum  of  several 
products  separated,  by  +  or  —  signs. 

Thus,  in  the  example  above,  the  given  expression  is  not  factored 
when  written  in  the  form  a(x  —  o  ?/)  +  2  b(x  —  5  ?/),  though  certain 
groups  of  its  terms  are  factored. 

Ex.  3.     Factor    a*  +  a^  +  a  +  1. 

We  have     a* -i- a^ -}- a -i- I  =  a^  (a  +  I)  +  (a  +  I)  Check,  a  =  2. 

=  (a2  +  l)(a  +1).     8  +  4  +  2  +  1  =  5-3 

15  =  15. 
The  student  should  group  the  terms  a*  and  a,  and  also  a^  and  1,  and 
obtain  the  same  result. 

16.  It  often  happens  that  an  expression  contains  groups  of  factors 
which  differ  only  in  sign. 

Ex.  4.    Factor  x^a  -  1)  -  ^/^(l  -  a). 

Since  1  —  a  =  —  (a  —  1),  we  may  transform  the  expression  so  that  both 
groups  of  terms  shall  contain  the  common  factor  a  —  1 ;  we  do  this  by 
writing 
z^(a  -  1)  -  2/2(1  -a)  =  x^(a  -  1)  +  y^(a  -  1)  Check. 

=  (x2  +  2/2)(a-l).  Letx  =  2,a  =  3,y  =  4. 

4  .  2  -  16  (-  2)  =  20  •  2 
40  =  40. 


190  FIRST  COURSE  IN  ALGEBRA 

Exercise  XII.    2 

Obtain  factors  of  the  following  expressions,  checking  all  results 
numerically  : 

1.  ac  +  ad+  be  +  bd.  8.  a"^  —  a^  +  a  —  1. 

2.  xi/  +  onv  —  yz  —  zw.  9.  ^*  +  4  ?/^  +  2  y  +  8. 

3.  bs  —  st  —  be  +  et  10.  m^  —  m  —  a  +  am. 

4.  m*  —  2  7W^  —  3  7rt  +  6.  n.  1  +  X  +  y  +  xy. 

5.  a'  —  la'^  —  Sa  -f  21.  12.  a^  +  abm  —  4cab  —  4^mb\ 
G.  2  w  +  3  7WW  4-  3  /?  4-  2  7W^.  13.  6  ar*  +  3  a-y  —  2  a«  —  ^j/. 

7.  ar'  — 4ic  +  a-y  — 4^^.  14.  3a*—  15a  +  106  — 2a*6. 

15.  oa;  +  a^  +  6aj  +  6y  +  ca;  +  cy. 

16.  by  —  b  —  ey  +  c  —  dy  +  c?. 

17.  wa;  —  my  +  mz  —  nx  +  ny  —  nz. 

18.  2mn  —  2ny  ^  mx  -\-  xy  +  2n^  —  nx. 

19.  216  — 5a  + 3a6- 2ac— 14c  — 35. 

20.  icz  +  x  — 5yz  — 5y  — Qz  — Q. 

21.  ax -\-  ay  —  bx  —  by  •\-  bz  —  az. 

22.  ax  —  bx  -\-  X  —  ay  +  by  —  y, 

23.  ca;  —  <?a;  +  cy  —  dy  +  C2;  —  (f^;. 

24.  aw  —  6;i  +  aw  —  6w  +  cw  +  cw. 

25.  a6a;  +  bex  +  caa;  +  a6y  +  bey  +  cay. 

26.  8  aa:  +  15  6y  +  12  6aj  4-  10  ay. 

27.  a^c  +  b\  +  aH  4-  6V  +  ah  +  6^c. 

28.  ax  4-  6y  4-  ay  +  62;  4-  a;:;  4-  6a;  +  ca;  4-  cy  4-  cz. 

29.  of  —  bg  -\-  ch  -{-  ag  —  bh  -\-  cf  +  ah  —  bf  ■\-  eg. 

30.  ar  +  6s  —  c^  4-  a5  4-  6^  —  cr  4-  a#  4-  6/-  —  C5. 

17.  Any  identity  obtained  by  the  process  of  multiplication  may, 
by  being  "read  backward,"  be  used  as  a  model  form  to  give  a 
factorization. 

Type  I.    Trinomial  Squares 

18.  From  Theorems  I  and  II,  Chapter  VII.  §§  18,  19,  we  obtain 

a^  ±  2  a&  4-  6'  =  (a  ±  h)\ 

The  type  expression  «^  ±  2  a6  4-  6^  consists  of  two  squares,  a^ 
and  6^,  having  positive  signs ;  the  remaining  term  ±  2  a6  is  double 


INTEGRAL  FACTORS  191 

the  product  of  the  numbers  a  and  6,  whose  squares  are  represented 
by  a^  and  IP: 

The  sign  of  the  middle  term  ±  2a6  is  positive  or  negative 
according  as  the  signs  of  a  and  h  are  the  same  or  different. 

19.  The  student  should  become  familiar  with  expressions  of  the 
type  a^  ±  2  a6  +  h^,  and  should  be  able,  when  any  two  of  the  three 
terms  a^,  ±  2  a6,  or  IP  are  given,  to  supply  the  third  term  which  is 
necessary  to  make  the  expression  a  trinomial  square. 

Ex.  1.  Supply  the  term  which  is  necessary  to  make  a^  +  (  )  +  25 
a  trinomial  square. 

Assuming  that  a^  and  25  are  the  first  and  third  terms  respectively  of  a 
trinomial  square  of  the  form  a^db  2aA  +  6^,  it  appears  that  the  missing 
middle  term  +  (  ),  represented  by  ±  2  a6,  may  be  obtained  by  finding  ± 
double  tlie  product  of  the  square  roots,  a  and  5,  of  the  first  and  third  terms 
a^  and  25  respectively  ;  that  is,  the  missing  term  is  ±  2  •  a  •  5  =  ±  1<^«- 

Since  the  missing  middle  term  is  assumed  to  be  positive,  the  required 
trinomial  square  is  a^  -f  10  a  +  25. 

Mental  Exercise  XII.    3 

Find  the  terms  which  if  supplied  will  make  the  following  expres- 
sions trinomial  squares  : 


1.  a^  +  (    )  +  9. 

17.  9t«^'-(     )  +  9. 

2.  c^  +  (    )  +  64. 

18.  16a'-(     )+  16. 

3.  c?2  +  (     )  +  100. 

19.  6^  +  (     )  +  c\ 

4.  ar^  -  (     )  4-  16. 

20.  a^  -  (     )  +  ^. 

5.  /  -  (     )  +  49. 

21.  4^- (     )^x^. 

^.  z^-{     )  +  81. 

22.  9A2-(     )+F. 

7.  <f  +  (     )  +  121. 

23.  a;2+  (     )  +  4/. 

8.  e^—  (     )  +  144. 

24.  /-(     )  +  64;.«. 

9.  4fl'  +  (     )+  1. 

25.    ;2'  -  (     )  +  81  w\ 

10.  9  6^  4-  (     )  +  1. 

26.    1  +  (     )  +  x\ 

11.   16  6-2+  (     )^  1^ 

27.    l-(     )+/. 

12.  se^/'-C   )  +  1. 

28.    1  -  (     )  +  169a^ 

13.  25/ -(     )+  1. 

29.    1  -  (     )  +  196  6^. 

14.  9aj^  +  (     )  +  4. 

30.    4a' +  (     )  +  9w^. 

15.   16/ +  (     )  +  25. 

31.  1  +  (  )  +  121  e. 

16.  49  52-(     )  +  4. 

32.    1  -  (     )  +  225  d\ 

192 


FIRST  COURSE   IN   ALGEBRA 


33.  1  +  (     )  +  256  k\ 

34.  16  6^+  (    )  +  25  c^ 

35.  ^d'-{    )  +  l^k\ 

25a;'+(     )  +  64/. 

49.y2+(     )  +  16/. 

36  tf 2  -  (     )  +  4  b''. 

64^^-(     )+i)c\ 

49ic''  +  (     )  +  49/. 


36. 
37. 
38. 
39. 
40. 


41.  81rt'2+  (     )_^  25  6=*. 

42.  121  a^*  -  (     )  +  Ub\ 

43.  9c?'»+  (     )  +  \(di)m\ 

44.  4/+  (    )  +  169;^=*. 

45.  Slc^-  (     )  +  \2\w\ 

46.  \mh^-  {     )  +  UqK 

47.  225  r'  -  (     )  +  36  v\ 

48.  25  s'  —  (     )  +  256 1\ 


49.    196»2+  (    )  +  25  r'*. 
Ex.    50.  «^+  2a6+  (     ). 

In  order  that  2  aft  shall  be  the  middle  term  of  a  trinomial  square  of  which 
one  of  the  terms  is  a^,  the  term  supplied  must  be  the  square  of  the  quotient 
obtained  by  dividing  2 ah  by  two  times  the  square  root,  a,  of  the  term  a^. 

From  2  «6  -^  2  a  we  obtain  6,  which  is  accordingly  the  term  whose  square 
fc2  is  require<l. 

Hence  the  desired  trinomial  square  is  a^  -f  2  oft  +  6^. 


51. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 
59. 
60. 
61. 
62. 
63. 
64. 
65. 
66. 
67. 
68. 

20. 

terms 


c2+2c</4-(    ). 


). 
)• 
)• 
). 
)• 
). 
). 
). 


a^H-  2ac+  ( 
6^  +  2  ^  +  ( 
c^  -  2  rA'  +  ( 
<^^+  4t/+  ( 
aj*+  6a;+  ( 
/+102/+( 
/-14y+( 

a^  —  2la+  ( 

4«2+  12a  +  ( 
9  6^  +  18  6  +  ( 
25c2+  40c  +  ( 
49c?2  — 28<^+  ( 
(     )  +  2w+  1. 

)  +  2i«+  1. 

)  + 42^+1. 

)  +  10;^+  1. 


). 


). 
)• 


69. 
70. 
71. 

72. 
73. 
74. 
75. 
76. 
77. 
78. 
79. 
80. 
81. 
82- 
83. 
84. 
85. 
86. 


( 
( 
( 

Before  attempting  to  factor  a  trinomial  in  which  two  of  the 
are  squares,  the  student  should  examine  the  middle  term  to 


+  16«^;+  1. 
+  ^a  +  a\ 

+  12  c  +  c\ 

-  14  c?  +  d\ 
+  2ak-\-  k\ 
+  2  mn  +  71^ 
-2ay  +  y\ 
-A:az  +  z\ 

-  10%  +  /. 
+  126  +  4^2. 
+  20c  +  4c^ 
-42«6  +  96^. 
+  96«y  +  64/. 
-90  6/?  +  81/^^. 

-  66  ca;  +  9  x^. 
+  120  aw  +  2b  w\ 

-  130%+  25/. 


INTEGRAL  FACTORS  193 

find  whether  or  not  it  satisfies  the  conditions  for  a  trinomial 
square. 

Ex.  1.    Factor  a:^  +  10  a;  +  25. 

Two  of  the  terms,  x^  and  25  are  positive  squares,  and  to  be  a  trinomial 
square  the  remaining  or  "  middle  term  "  should  be  double  the  product  of 
the  square  roots  of  these  terms,  x  and  5,  that  is  2(a:)(5)  =  10  ar. 
Since  the  middle  term  is  10  a:,  we  may  write 

Check.    Let  a:  =  2. 
ar2  +  lOx  +  25  =  (a:  +  5)2.  4  +  20  +  25  =  7^ 

49  =  49. 
Ex.  2.    Factor  49  ar^  _  28  ary  +  4  y\ 

We  find  that  the  terms  49x2  and  4i/2are  the  squares  of  Tx  and  of  2y 
respectively.  The  remaining  term  —  28  xy  is  double  the  product  of  these 
terms  7x  and  —  2y. 

Hence  we  have  Check.     Let  x  —  %y  —  ^. 

49a;2_28a;2/  +  4i/=(7a;-2  2/)2.         196-168  +  36=    8^ 

64  =  64. 
Ex.  3.   Factor  72  a:»i/«  _  81  a;*  -  16 1/«. 

The  expression  should  be  placed  in  "minus  parentheses,"  because  the 
squares  81a:*  and  16y*  should  appear  as  positive  numbers  to  satisfy  the 
type  expression  a^  ±  2  ah  +  ly^. 
Accordingly,     72  a:*/  -  81  x«  -  16  /  =  -  [81  a:«  -  72  x^f  +  16 !/«] 

=  -  [(9  a;8)2  -  72  x^f  +  (4  y^Y\ 
=  _(9x«-4i/«)2. 

Check.  x  —  %y—\. 
576  -  5184  -  16  =  -  (72  -  4)2 
_  4624  =  -  4624. 

Exercise  XII.    4 

Factor  the  following  expressions,  verifying  all  results  by  numerical 
checks  : 

8.  y'-\-^h^  16. 

9.  c2+  12c  +  36. 

10.  6?=*-  18  c? +  81. 

11.  25  +  \^a-\-a\ 

12.  49-  14?/  +  y. 

13.  64-  16;^  +  ;^'. 
14  100  +  20^ +  ^"^ 

13 


1. 

a;2+  2ic+  1. 

2. 

?w2-2w+  1. 

3. 

z^  —  1zw->r  w\ 

4. 

a''-^2ak  +  k\ 

5. 

r^  —  2rv  +  v\ 

6. 

q^-2qx-\-  x\ 

7. 

a"  ■\-  \a-\-  4. 

194  FIRST  COURSE   IN  ALGEBRA 

15.  4a' +  4a  +  1.  32.  2566'^+  96  c^  +  9cP. 

16.  9  ^'  +  6  ?>  +  1.  33.  64  a%^  +  80  a6c  +  25  c^ 

17.  36  c^  +  12  c  +  1.  34.  16  a;*  -  56  xijz  +  49  y^z\ 

18.  81  w''—  18w+  1.  35.  4c'»—  36(^«  +  81c?V. 

19.  16  a'^  +  24  aft  +  9  h\  36.  25  a'^ft'  +  40  abed  +  16  cW 

20.  9  c'  +  30  cfl?  +  25  cT*.  37.  36  a V  —  84  aftar^  +  49  by. 

21.  25  F  +  70X-ir  +  49  w\  38.  9  by  +  48  ftcys  +  64  c'z'. 

22.  121  5'^  +  44  5^  4-  4 1\  39.  49  ay  +  42  ahxy  +  9  ftV. 

23.  49  A^  -  140  ^^  +  100  /.  40.  4  ah""  -  20  acmz  H-  25  c2»2^ 

24.  16  g^  +  72  ^w  4-  81  w^  41.  25  a'ftV  -  60  abed  +  36  <^. 

25.  121  k^  -  242  ^.<?  +  121  5*.  42.  81  a;*  +  180  xijzw  +  100  fzV. 

26.  36  m-  +  108  ww  +  81  ?i\  43.  (a;  +  3^)2  +  2  (a;  +  3/)  +  1. 

27.  81  ^  -  126  de  +  49  e".  44.  (a  -  bf  -  2  (a  -  b)  +  1. 

28.  64^2  +  320  ^m;  +  400  m;^  45.  (m  +  n^  +  6  (w  +  w)  +  9. 

29.  225  r*  —  120  rs  +  16  s^.  46.  [x  -  yf  +  io(x  —  y)  +  25. 

30.  169  f  -  260  ^m;  +  100  w\  47.  16  +  8  (a  +  6)  +  (a  +  6)*. 

31.  100  a''  -  280  a6  +  196  ft'*.  48.  36  -  12  (a  -  a;)  +  (a  -  a;)^ 

49.  (a  +  ft)«  +  2(a  +  ft)(a;  +  y)  +  (a:  +  yf. 

50.  (a  -  ft)2  -  2  (a  -  ft)(a;  -  y) -\-  (x  -  yy. 

51.  (ft  +  c)'  -  8  (ft  +  c)(x  +  y)  +  16  (a;  +  yy. 

52.  9  (a  +  ft)'  +  12  (a  +  ft)(c  +  </)  +  4  (c  +  <?)'. 

53.  49  (d  +  ky  —70(d  +  k)(m  -\-  w)  +  25  (m  +  wy. 

21.  If  an  expression  has  the  form  of  a  trinomial  square, 
a'  ±  2  aft  4-  ft',  we  shall  call  the  "middle  term,"  represented  by 
±  2  aft,  the  finder  term. 

The  finder  term,  ±  2  aft,  of  an  integral  rational  trinomial  square 
a'  ±  2  aft  +  ft',  contains  as  factors  the  terms  a  and  ±  ft  of  the  bi- 
nomial a  ±  ft  of  which  the  trinomial  is  the  square. 

We  may  often,  by  using  some  particular  term  as  a  finder  term, 
select  from  a  set  of  four  or  more  terms  a  group  of  three  terms,  which 
takan  together  form  a  trinomial  square. 

Ex.  1.    Factor  a:^  +  ?/'  -  2  a:  +  2  a:?/  +  1  -  2  y. 

Since  there  are  three  separate  squares,  cc^,  y'^  and  1,  and  three  terms,  2  ary, 
—  2  a;  and  —  2  1/,  each  having  a  coefficient  2,  we  are  led  to  suspect  the  exist- 
ence of  one  or  more  trinomial  squares  among  the  terms  of  the  given  ex- 
pression. 


INTEGRAL  FACTORS  195 

"We  may  group  x'\  y^,  and  2  xy,  since  the  product  of  the  square  roots  x 
and  y  of  x^  and  y^  is  doubled  to  form  the  "  finder  term  "  2  xy. 
We  may  write, 

x^-\-y^  -2x  +  2xy  +  l  -2y  =  x^  +  2xy  ^y"^  -2x  -2y  +  l 

=  (x-hyy-2x-2y  +  l. 
Since  (x  -\-  y)^  and  1  are  squares,  we  may  look  for  tioice  the  product  of 
their  square  roots,  (x  +  y')  and  1,  that  is,  for  2{x  -\-  y). 

This  product  is  obtained  by  writing  —  2  a;  —  2  i/  in  the  form  —  2(x  -{-  y). 
Accordingly   (x  -\-  yY  —  2x  —  2y  +  I    may   be   written   as   (x  +  y}'^  — 
2  (x  +  i/)  +  1,  which  is  the  square  of  x  -\-  y  —  1.  Check. 

Hence,  x^-\-y^  —  2x  +  2xy  +  l—2y  =  (x  +  y  —  iy.         Let  a:  =  2,  y  =  3. 

4  +  9-4+  12 +1-6=  (2 +  3-  1)2. 
16=  16. 
We  may  obtain  the  same  result  by  grouping  either  x^,  —  2x  and  1,  or 
y%  —  2y  and  1 . 

That  hx^  +  v^-2x  +  9^y+i-2v  =  \  '''^''  ^"^  -  ^Y  +  ^  0)  +  0)' 
ltiati8,ar   +y        Zx  +  2xy-\-l       22/_|      ^^     (^  _  i)2  +  2  (?)  +  (])2 

=  (x  +  y-l)2. 
The  student  should  carry  out  the  solutions  as  suggested  above. 
Ex.  2.    Factor  m^  +  2 mu  -\-u^  +  2mz  +  2 nz  +  z^. 

The  terms  may  be  grouped  in  any  one  of  the  ways  (1),  (2),  or  (3)  as 
suggested  below : 

{  (1)  7/i2  +  2wm  +  n2withthecor-  (  (1)  +  2(m  +  n)2! leaving  (  (\)+zK 
^  (2)  m2  +  2ms+22   responding    -^  (2)  +  2(w  +  «)w     the     ■}(2)+rA 
(  (3)  71^  +  2  wxj  +  z^      groups         (  (3)  +  2(u-{-z)m  squares  (  (3)  +  m^. 
Of  these  three  possible  arrangements,  we  will  use  the  first, 
(1)  m^  +  2mn  +  n^  +  2(w  +  n)z  +  z^=  (m  +  n)^  +  2(m  +  n)z  +  z^ 

=  [(m  +  n)  +  2^         Check. 
=  [?7i  +  n  +  zf.  Let  m  =  2, 

?i  =  3,  2!  =  1. 
4  + 12 +  9  +  4  +  6+1  =  (2 +  3 +1)2 
36  =  36. 
The  student  should  show  that  the  adoption  of  either  of  the  arrangements 
of  the  terms,  (2)  or  (3),  leads  to  the  same  result. 

Exercise  XIL     5 

Obtain  factors  of  the  following  expressions,  checking  all  results 
numerically  : 

1.    a^  +  2ab  +  h''  +  2a  +  2b  +  I. 

10  m  +  10/2  +  25. 


196  FIRST  COURSE  IN  ALGEBRA 

3.  a^  +  b^  +  x"  -\-  2ab  +  2ax  +  26a;. 

4.  ar^+ 6a;^+ 9/+ 2a;+ 6y  +  1. 

5.  c^  +  d^  -{-  e^  -  2cd  —  2c€  +  2de, 

6.  c'^  +  cT*  +  4  —  2  cfi?  +  4 c  -  4(/. 

7.  ar^  +  3/'  +  i;=^+ 22a;  — 2;jy-2a:iy. 

8.  4a^  + 6^+  1  +  4tf6  + 4a  + 26. 

9.  9a*+ 46*  +  c''- 12a6-6ac  +  46c. 

10.  a{a  +  2X')  +  w(m;  +  2a)  +  A*(X^  +  2w), 

11.  6(6  +  2  c)  +  c(c  +  2  (/)  +  c?(fl?  +  2  6). 

12.  a{a  —  2  6)  +  6(6  +  2  c)  +  c(c  —  2  a). 

13.  {x"  +  2yc)  +  (/  +  2a;y)  +  {z''  +  2a;2). 

14.  a^  +  }/+  c"-  2(ah  -  ac  +  be). 

15.  ^it^-\-y^-\-^2^+2{2xy -(ixz-^yz). 

Type  II.    The  Difference  of  Two  Squares 
a;' -2/2 

22.   From  the  converse  of  the  identity  in  Chap.  VII.  §  20,  we  have 

a^  —  y'^  =  {x-{-y){x—  y). 

0r,  the  difference  of  the  squares  of  two  numbers  may  be  written 
as  the  product  of  the  sum  and  difference  of  the  numbers. 

Ex.  1.   Factor  64  a:2  -  9. 

Since  64  x^  is  the  square  of  8  ar,  and  9  is  the  square  of  3,  we  may  write  as 
factors  of  64  a:^  —  9  the  product  of  the  sum  8  x  +  3  and  the  difference  8  a;  —  3 
of  these  same  numbers. 

Check.     Let  a:  =  2. 
That  is,  64a:2  -  9  =  (8  ar  +  3)(8a:  _  3).  256  -  9  =  19  .  13 

247  =  247. 
Ex.  2.    Factor  lOOa^ftV  _  25  dK 

The  terms  of  the  binomial  quotient  4  a*h\^  —  d^y  obtained  by  dividing 
the  terms  of  the  given  expression  by  the  common  numerical  factor  25,  are 
squares. 

Hence,  the  binomial   difference,  4  a*6 V  —  d"^,  may  be  expressed  as  the 
product  of  the  sum  2  a%(^  +  d,  multiplied  by  the  difference  2  a%c^  —  d. 
Hence,  we  have  100  a^ly^c*  _  25  c?2  =  25  [4  a*hh^  -  d^ 

=  25  (2  a%c^  +  d)(2  a%c^  -  d). 
Check.     Let  a  =  3,  6  =  2,  c  =  I,  fZ  =  4. 
32000  =  32000. 


INTEGRAL   FACTORS  197 

Mental  Exercise  XII.    6 
Factor  each  of  the  following  expressions  : 

1.  a"  —  4.  29.  289  « V  _  49  ^y. 

2.  ^2  -  16.  30.  324a27i=^  -  64  cW. 

3.  c'  --  25.  31.  121  a;'  -  1. 

4.  ^2_49.  32.  64 a;**- 9. 

5.  ^"-81.  33.  25/- 16. 

6.  25 -a;'*.  34.  36si°-49. 

7.  36-?/'^.  35.  64F=^- 100. 

8.  64-;^=^.  36.  81 -121  a". 

9.  81a''-  1.  37.  1446^- 121c*. 

10.  49Z»2-1.  38.  Ua'-UhK 

1 1.  100  a"  -  b\  39.  49  x^  -  36  t/^^ 

12.  4  c'^  -  25  d\  40.  9  a'''  -  25. 

13.  9  7^—  162^^.  41.  a;^"*  —  64/'". 
14  16  ^'^  -  49  t\  42.  49  a"'  -  100  U^K 

15.  25  ^2  -  64  m\  43.  180  a**"  -  121 7i^. 

16.  1  —  121  n\  44.  64  a^*"  —  49  r\ 

17.  1-169  P.  45.  a;^^-  1. 

18.  144  ^2  -  25  f.  46.  4  «2/^V  -  1. 

19.  9  A*  -  100  q".  47.  9  a;y;j2  -  1. 

20.  64  a^h^  —  25.  48.  16  a^^^c^  —  (P. 

21.  49  a;y  -  64.  49.  25  a^%'c^  -  d^e\ 

22.  81  -  36  mV.  50.  twV"  -  1. 

23.  36  -  81  c't^.  51.  rV  -  if''. 

24.  25 p^  -  144  a;V:  52.  a^*  —  b^''. 

25.  100  a''^;*  -  81  c^d\  53.  4  x'''  -  1. 

26.  196  ar*/  -  100  2W.  54.  9  a""^  -  ^>^ 

27.  169  aV  -  225  6W  55.  16  a;'"  -  f. 

28.  256  a^d^  -  144  ^>V.  56.  4  a*"  -  9. 

23.   The  difference  of  the  squares  of  either  binomials  or  polynomials 
may  be  factored  by  applying  the  method  under  consideration. 


198  FIRST  COURSE  IN  ALGEBRA 

Ex.  1.    Factor  (a  +  b)-  -  (c  +  d^. 
We  have 

(a  +  by  -  (c  +  dy  =  [{a  +  h)  +  (c  +  d)][(a  +  ?>)  -  (c  +  (^)] 
=  [a  +  6  +  c  +  rf][a  +  6  -  c  -  d]. 

Check.     Let  a=zb  =  3,  c  =  d  =  2. 
36  -  16  =  10  X  2. 
20  =  20. 
Ex.  2.   Factor  (x  -  yY  -  (m  -  n)^. 

We  have 

(x  _  yy  -  {in  -  n)2  =  [(x  -  y)  +  (m  -  ?i)][(x  -  i/)  _  (m  -  ?i)] 
=  [x  —  ?/  +  wi  —  ?j]  [x  —  y  —  m  +  w]. 

Check.     Let  x  =  4,  y  =  2,  w  =  3,  w  =  2. 
22  _  12  _  3  .  1 

3  =  3. 
Exercise  XII.    7 

Obtain  factors  of  the  following  expressions,  checking  all  results 
numerically : 

1.  {a  +  hf  -  1.  15.  49  (a'  +  hf  -  144  (a  +  b^. 

2.  (a  +  ^)'  -  4  c^  16.  16  \a^  +  ^^*  -  121  (a  +  6)*. 

3.  \x-y\-  zf  -  1.  17.  64  (a  +  ^>  +  c)^  -  169  ^'. 

4.  (a  +  6  — c)^  — 9^.  18.  121  (a  — ^>  + c)'— 225^2. 

5.  {a  +  by—{x-y-^zy.  19.  25(a-^  +  c)^-81(ic-^-;^)2. 

6.  (a  -  6)2  -  (a;  +  ^  -  zy.  20.  100  icV-*  -  196  «'  (6  +  c)^. 

7.  (x-y-  zy  -  (a  +  6)^.  21.   121  a^  {b  +  c)^  -  81. 

8.  {a  +  b  —  cy-^d{x-yy.  22.  im  b^  {c  +  d)""  -  l^&  e\ 

9.  9  (a  +  ^)^  —  16  (^  +  w)"-  23.   196  ic^  (y  —  ;s)'^  —  225  «/A 

10.  9  (a  -  6)^  -  49  (c  -  <^)2.  24.  36c*''((/-^)2- 121er2(7/  +  ;^)'. 

11.  4ar'  —  49  (a  +  6  +  c)^.  25.  lUa^(a  +  b  +  cy  —  225  c?^. 

12.  64  0  +  ^)2  -  121  (k  +  r)2.  26.  64  ar^  (w  -  ny  -  225  P^^ 

13.  SQ(g  —  xy-Ud(h-yy.  27.  81a2(m+/j)'-256^/Xa^+^-^)2. 

14.  225  (a  +  by-  225  (c  +  i)^. 

24.  Many  polynomial  expressions  may  be  so  transformed,  by 
suitably  grouping  the  terms,  as  to  appear  as  the  difference  of  two 
squares. 

Ex.  1.    Factor  a^  +  2ah  +  h^  -  c^  +  2  cd  -  d^. 

The  terms  may  be  so  grouped  that  the  expression  will  appear  as  the 
difference  of  two  trinomial  squares,  as  follows  : 


INTEGRAL  FACTORS  199 

a«  +  2  a6  +  62  _  0-2  +  2  cd  -  d^  =  {0^  +  2ah  -{-h'^)  -  {c^  -  ^  cd  +  ^2) 

=  (a  +  6)2  -  {c  -  dy 
=  [((t  +  h)  +  {c  -  d)][(a  +  b)  -{c  -  d)] 
=  [a  +  6  +  c  -  rf]  [ft  +  6  -  c  +  rf]. 

Check.     Let  ft  =  4,  6  =  3,  c  =  2,  c?  =  1. 
16 +  24 +  9-4  + 4 -1  =  8-6 
48  =  48. 
Exercise  XII.    8 

Factor  the  following  expressions,  checking  all  results  numerically  ; 

1.  x''  +  2xi/  +  f-  z\  13.   1  -  a^  _  2  ab  -  b\ 

2.  m^  -\-  2  mil  +  ii^  —  w^.  14.   1  —  x^  —  2x}j  —  ?/. 

3.  a'-2a6  + ^'-/.        '     15.  4  -  c^  -  2  c^  -  i^. 

4.  c'-2cd-\-  (P  —  k\  16.  ^-m^  ->r2mn  —  n\ 

5.  a'  +  2ab-{-  b^  -  4..  17.  25a-'-  10a/>  +  6' -  s^. 

6.  c'^  —  2  cc?  +  fl?2  -  9.  18.  c^  —  18  c  +  81  -  (P, 

7.  a^  +  2a6  +  ^/^^  -  1.  19.  A'  -  20A  +  100  -  F. 

8.  /  — 2^A:  +  F  — 4wl  20.  16a^  —  8«6  +  6^  -  16. 

9.  P  -2kx  +  X''-  16/.  21.   49c?2  -  Udr+r""  -  49. 

10.  a^+  6a6  +  9  ^'^  -  c''.  22.  81^  -  36/?a;  +  4a;2  -  1. 

11.  cc'-lOa!^    -\-25f  —  z^     23.   1  —  9^2  —  30a^  -  25^'. 

12.  w*  -  12  w?«  +  36  ?*'  —  ^''.    24.  49  a^  —  28  a6  +  4  6'  —  9  c^. 

25.  36  ^  +  60  fl?^  +  25  -s'^  —  121  w\ 

26.  64c2-48c/i  +  9;*''- 16t;^ 

27.  a'b''  +  28  a6  +  196  -  169  c\ 

28.  cY  —  24  c^  +  144  -  121  F. 

29.  a^-i-2ab  +  b^-c^-2cd-  d\ 

30.  c*  -  2  cA  +  A'  - 

31.  F+  2/:w  +  w* 

32.  h^  —  2kt  +  t'- 

33.  9a''  +  6a^  +  ^2  -  6-2  -  led  -Ad^ 

34.  a^  -  lOaiij/  +  25/  —  36;52  —  122;^^;  -  w\ 

35.  a''  -  14a^  +  49  6"  -  a;^  —  Uxi/  —  64/. 

36.  100^/2- 20^>G?  +  c?'-/+  183/^-812;^ 

37.  48  a'  —  3  b\  41.  ax^'t/  —  axi/. 

38.  7  a;'  -  63  f.  42.  a;'  -  /  +  a;;?  +  ?/;2^. 

39.  45  -  125  a'6^  43.  x""  —  f  +  x  —  1/, 

40.  28  a^y  -  7  (c  +  c^)''.  44-  a'6  -  ab^  +  «'/>  +  a6^ 


m' 

—  2mr- 

-r\ 

-r^ 

+  2r5- 

•s\ 

n" 

+  2nij- 

■f' 

200  FIRST  COURSE   IN  ALGEBRA 

25.  An  expression  of  the  form  x*  ±  hx-y^  +  y/*,  where  h  has 
such  a  value  that  the  trinomial  is  not  a  square,  may  be  transformed 
into  one  by  adding  a  square  represented  by  k'^x^y^^  and  combining 
the  term  thus  introduced  with  ±  kxhj\ 

By  subtracting  kVi/\  the  value  of  the  transformed  expression 
will  become  equal  to  that  of  the  given  expression,  but  will  appear  as 
the  difference  of  two  squares,  in  which  form  it  may  be  factored. 

26.  The  value  of  the  term  Px^y\  used  in  making  the  transforma- 
tion, is  obtained  by  subtracting  the  given  term  ±  hx^y^  of  the  ex- 
pression ic*  ±  hx^y"^  -f  y,  which  is  not  a  trinomial  square,  from  the 
required  middle  term  of  the  trinomial  square  of  which  aj*  and  y^  are 
the  first  and  third  terms  respectively. 

Ex.  1.    Factor  ar*  -f-  9  a:^  +  25. 

If  x*  and  25  are  the  first  and  third  terms  of  a  trinomial  square,  the 
middle  term  must  be  ±  10  x*. 

Subtracting  the  given  middle  term,  9x*,  of  the  expression  x*  -|-  Ox'*  +  25 
which  is  not  a  trinomial  square,  from  the  required  middle  term  ±  10  x^  of 
the  required  trinomial  square  x*  +  10  x^  -}-  25,  we  obtain  x*  and  also  —  1 9  x**. 

Using  the  square  x*,  we  may  transform  the  given  expression  and  obtain 
its  factors  as  follows  : 

X* +  9x2  + 25  =  x*  + 10x2  +  25 -x3        Check.     Let  x  =  1. 
=  (x2  +  5)2  -  x2  1  +  9  +  25  =  7  •  5 

=  (x2  +  5  +  x)(x2  +  5-x).  35  =  35. 

If,  by  adding  and  subtracting  —  19  x^,  the  given  trinomial  had  been  so 
transformed  as  to  have  a  negative  middle  term,  —  10  a;^^  the  transformed 
expression  would  have  appeared  as  a  sum  (x*  +  5)^  +  19x2,  and  in  this 
form  would  have  had  no  "  real "  factors. 

Ex.  2.    Factor  9x*  -  l^xhf'^  +  16 j/*. 

A  trinomial  square  of  which  9x*  and  16 1/*  are  the  first  and  third  terms 
must  have  for  a  middle  term  ±  2(3  x2)(4  y"^)  =  ±  24  x^y^. 

Subtracting  the  given  middle  term  —12x^y^  from  the  required  middle 
term   ±  24x2j/2,  we  obtain  36x2|/2,  and  also  -  12x2^2, 

Using  the  square,  36  x^y^j  we  may  transform  the  given  expression  and 
obtain  its  factors  as  follows  : 

9x*  -  12  xV  +  167/  =  9x^  +  24x2|/2  +  I6i/*  -  36x2i/2  Check. 

=  (3x2  +  4y2)2  _  (6arj/)2  x  =  y=l. 

=  [3x2  ^  4y'z  +  6xi/][3x2  +  42/2  -  6xi/].     13=  13. 

The  student  should  explain  the  reason  for  not  using  —  24x2j/2  as  a  "mid- 
dle term  "  in  the  transformed  trinomial  above. 


INTEGRAL  FACTORS  201 

Exercise  XII.     9 
Factor  the  following  expressions,  checking  all  results  numerically: 

1.  X*'  -f  ar^/  +  /.  13.  4  ic*  —  8  xhf  +  y^ 

2.  x^  +  xY  +  /.  14.  9a;*  +  3£cy  +  4^^^ 

3.  a;*  +  ic"  +  1.  15.  9  ^^*  -  21  F  +  4. 

4.  a*  —  7  a^  +  1.  16.  100  a*  —  49  a^  +  4. 

5.  a;*+  3ar^+  36.  17.  36c*-40c2+  9. 

6.  a*—  14«2+  1.  18.  36m*+  116  7wV+  121  w*. 

7.  rt*  +  9  a"  +  25.  19.  25  +  9  m*  +  26  m\ 

8.  4a;*—  13ar»+  1.  20.  25a*  +  101  rt'6' +  1216*. 

9.  25a*-  lla=^+  1.  21.  49a;*+  64?/*+  87  a;y. 

10.  a;*  —  13  xSf  +  A.y\  22.  25  w*  +  36  w*  +  35  m^n\ 

11.  a*  +  24  a'b''  +  196  ^>*.  23.  25  a«  +  81  6»  +  41  a*6*. 

12.  4  a*  +  16  a^  +  25.  24.  ix^^  —  34  a^y  +  81  /'. 

TYPE  III.    Trinomials  of  the  Type  ic'^  +  sx  +  p 

27.  From  the  distributive  law  for  multiplication,  we  have 

{x  +  a)(x  +  6)  =  ar*  +  (a  +  b)x  +  ah. 

By  comparing  this  result  with  the  type  expression,  it  appears 
that  the  coefficient,  a  +  6,  of  a;  corresponds  to  the  coefficient  s  in 
the  t)rpe  expression,  and  the  constant  term  ab  corresponds  to  the 
term  p. 

Hence,  reading  the  identity  backward,  the  necessary  and  suffi- 
cient conditions  that  the  trinomial  x^  -^  sx  +  p  may  be  factored, 
are  that  the  first  term  should  be  a  positive  square  with  the  coeffi- 
cient 1,  and  that  the  coefficient,  5,  of  x  should  be  the  sum  of  two 
numbers  whose  product  is  the  constant  term  p. 

28.  To  factor  trinomials  of  the  type  x^  +  sx  +  jo,  examine  first 
the  product  term  p^  and  separating  it  into  pairs  of  factors^  select 
that  pair  whose  sum  is  the  coefficient^  s,  of  x. 

Then  write  two  binomial  factors,  each  having  x  for  a  first  term^ 
and  for  a  second  term  erne  of  the  factors  of  the  pair  whose  sum  is  s. 

Ex.  1.    Factor  a:^  +  7  a;  +  12. 

This  expression  is  of  the  type  x^  +  sx  -{■  p,  because  the  first  term,  x\  is 
positive,  with  the  coefficient  1  ;  the  second  term,  7  x,  contains  x  to  the  first 
power;  and  the  third  term,  12,  is  free  from  x. 


202  FIRST  COURSE   IN   ALGEBRA 

We  find  that  the  only  possible  pairs  of  integral  factors  of  12,  whose 
product  is  12,  are  the  numbers  in  vertical  columns  : 

(12,     6,     4. 

1    1,     2,     3. 

7. 

The  sum  of  the  factors  4  and  3  is  7,  which  is  the  coefficient  of  x  in  the 

term  7  x.  Check,     x  —  %     4+ 14  4-12  =  6-5 

Hence,  we  may  write,  a;^  +  7  x  +  12  =  (x  +  4)  (x  +  3).  30  =  30. 

29.  Whenever  the  double  signs  ±  and  =F  (read  "  plus  or  minus  " 
and  "  minus  or  plus  "  respectively)  are  used  in  algebra  before  num- 
bers which  are  not  separated  by  an  equality  sign,  it  is  agreed  that, 
■whenever  we  take  the  upper  or  lower  sign  of  either  double  sign,  we 
must  take  the  corresponding  sign  of  the  other  double  sign.  (See 
also  Chap.  XXII,  §  12.) 

We  must  never  take  the  upper  sign  of  one  double  sign  and  the 

lower  sign  of  the  other  in  the  same  calculation. 

T.         ^      „  _  ^  <  either    5  +  3-2  =  6, 

E.g.   5  ±3^2  means-!  ^^„      ^      / 

^  (  or  5-3  +  2  =  4. 

_,  ,  .^  f  neither  5  +  3  +  2  =  10, 

But  It  means  ^  k      o      «      a 

(        nor  5  —  3  —  2  =  0. 

Ex.  2.    Factor  a:«  +  8  x  -  20. 

Comparing  with  the  type  expression  x^  +  sx  +  p,  it  appears  that  x^  +  8  x 

—  20  satisfies  the  conditions  for  form,  and  that  of  any  pair  of  factors  of  —  20 
whose  product  is  —  20,  one  factor  must  be  positive  and  the  other  negative. 

As  possible  pairs  of  integral  factors  of  —  20,  we  have  : 

J  ±  20,  ±  10,  ±  5. 

IT    1,T    2, +  4. 
±    8 
The  coefl&cient,  8,  of  x,  of  the  term  8  x,  is  the  sum  of  the  factors  10  and 

—  2,  which  are  found  in  the  second  column.  Check,     x  =  3. 
Hence  we  may  write,  x*  +  8  x  -  20  =  (x  +  10)  (x  -  2).     9  +  24  -  20  =  13  •  1 

13=13. 

Ex.  3.    Factor  x^  -  9  x  +  14. 

The  expression  x^  —  9  x  +  14  is  of  the  type  x^  +  sx  +  p,  on  condition 
that  —  9  is  represented  by  s  and  14  by  j?. 

Since  the  product  term,  14,  is  positive,  the  signs  of  the  factors  of  the  pair 
whose  product  is  14  and  whose  sum  is  —  9,  must  be  like,  and  to  produce 
the  sum  —  9,  these  numbers  must  both  be  negative. 


INTEGRAL   FACTORS  203 

Hence,  the  possible  pairs  of  factors  of  14,  whose  product  is  14,  are 
5-14,     -7. 
}-    1,     -2. 
-9. 

The  sum  of  the  factors  —  7  and  —  2  is  —  9,  which  is  the  coefficient  of  x 
in  the  term  —  9  a;.  Check,     x  =:  3. 

Hence  we  may  write,  x^  —  9x+]4  =  (x  —  7)(z  —  2).     9  —  27  +  14  =  —  4  •  1 

-  4  =  -  4. 
Ex.  4.    Factor  (a  +  6)2  +  5  («  +  6)  +  6. 

This  expression  is  of  the  type  x^  +  sa:  +  p,  on  condition  that  the  binomial 
a  +  6  in  the  given  expression  is  represented  by  x  in  the  type  expression. 
Hence,  we  have, 

(a  +  6)2  +  5  (a  +  6)  +  6  =  [{a  +  6)  +  2][(a  +  6)  +  2] 
=  (a  +  6  +  3)(a  +  6  +  2). 

Check. 

Let  a  =  6  =  2. 

42  =  42. 

30.  If  the  sign  of  the  coefficient  of  the  highest  power  of  the  letter 
with  respect  to  which  a  given  expression  is  to  be  factored  is  nega- 
tive, we  may,  by  placing  the  given  expression  in  minus  parentheses, 
so  transform  it  as  to  obtain  an  expression  in  which  the  coefficient  of 
the  highest  power  is  positive. 

In  this  form  it  may  be  treated  by  the  methods  for  factoring  ex- 
pressions of  the  form  a^^  +  so:  +  p. 

Ex.  5.   Factor  12  +  4  a:  -  x*. 

We  have         12  +  4a;  -  a:^  =  -  [x^  -  4a:  -  12] 

=  -  (x  -  6)(x  +  2)  Check.     Let  x  =  2. 

=  (6-x)(2  +  x).  12  +  8-4  =  4.4 

16  =  16. 

Exercise  XII.     10 

Obtain  factors  of  the  following  expressions,  checking  all  results 
numerically  : 

1.  a^+3a  +  2.  6.  x'^  nx+  18. 

2.  x'^+9x+  14.  7.  a^  —  12a;  +  20. 

3.  a^'+la-h  10.  8.  a^  -  15a  +  26. 

4.  x^  +  Sx+  15.  d.  i^+  Ix-  18. 

5.  x''  +  lOic  +  24.  10.  a^+  V2a-  45. 


204  FIRST  COURSE  IN   AI.GEBRA 

11.  ar»4-  5a;— 24.  32.  c^H-  22c+  112. 

12.  m^-^  dm +  8.  33.  ar*  +  14a;  +  13. 

13.  a'*  +  8a -9.  34.  a;^  +  16a;  +  55. 

14.  ar*  —  9a;  +  20.  35.  f  +  Idi/  +  60. 

36.  z^+  32  z+  112. 

37.  (P-  13  c? -68. 

38.  s^+  115-  180. 

39.  w^  +S5W+  300. 

40.  ny^  -}-  5w  —  300. 

41.  c^-  15  c -250. 

42.  «*^2  +  dab  +  20. 

43.  mH'^  —  A.mn-  140. 

44.  6V+.26  6C  +  165. 

45.  c^/  +  36  aj  +  320. 

46.  a''k^-\-  39aA:  +  380. 

47.  a'm'^  —  3  am  -  340. 

48.  (x  +  y)''  +  7  (a;  +  3/)  +  10. 

49.  {x-\-yy+n{x  +  y)  +  28. 

50.  (a-6)2  +  13  (a  -  ^)  +  42. 

51.  {x-yy+{x-y)-12. 

52.  {a~hy-l(a-h)-  18. 

31.    Certain  expressions  of  the  form  x^  +  saf*  +  ^,  containing 
only  two  powers  of  a  particular  letter,  and  one  of  the  powers,  a;^"*, 
being  the  square  of  the  other,  a;"*,  may  be  written  in  the  form 
(a.-'")'^  +  ^(af )  +  p,  and  factored. 
Ex.  1.    Factor  ar^o  +6x5  +  8. 
Since  x^^  is  the  square  of  x^,  we  may  write, 

a;io  ^  6  z^  +  8  =  {y^y  +  Q{x^)  +  8  Check.     Let  a:  =  1 . 

=  [x^  4-  4)(a;5  +  2).  1  +  6  +  8  =  5-3 

15  =  15. 

Ex.  2.    Factor  x«  +  6x8  -  27. 

Observe  that  x*  is  the  square  of  x*,  and  that  the  factors  9  and  —  3  of  —  27 
produce  by  addition  the  coeflBcient,  +  6,  of  x^. 

Hence  we  may  write,  Check.  Let  x  =  2. 

x«  +  6x«-27  =  (x3  +  9)(x8-3).  64  +  48  -  27  =  17  •  5 

85  =  85. 


15. 

a;^+  19a;  +  18. 

16. 

m^  —  m-  110. 

17. 

c^  +  10c  -  39. 

18. 

ar'—  11. T-  12. 

19. 

a*-23rt  +  60. 

20. 

a''-  18  a +  32. 

21. 

7??^  +  3  m  —  54. 

22. 

;w^+  15  7W  —  34. 

23. 

a*+  17  a +  42. 

24. 

0"+  18a  +  77. 

25. 

a*+  23a +  90. 

26. 

a;^  — 25a;+  100. 

27. 

c*  +  50  c  +  600. 

28. 

a^+  17a -38. 

29. 

c*-28c+  75. 

30. 

a*  — 26a  +  88. 

31. 

c''  +  44  c  —  45. 

INTEGRAL  FACTORS  205 

Ex.  3.    Factor  25  m^  +  15m  +  2. 

Observe  that  25  m^  =  (5  m)^,  and  that  we  may  write 

25m2  4-15m  +  2  =  (5m)2  +  3(5m)  +2. 
The  expression    (5  rri^)  +  3  (5  w)  +  2  is  of  the  type  x^  +  sx  +  p ;  5  m 
corresponding  to  x,  3  to  s,  and  2  to  the  constant  term  p. 

Hence,  since  the  sum  of  the  factors  2  and  1  of  the  constant  term  2,  is  3, 
we  may  write 

25  m^  +  15  m  4-  2  =  (5  m)^  +  3(5  m)  +  2 

=  (5  m  +  2)(5m  +  l).     Check.     Let  m  =  2. 
100+30  +  2  =  12  •  11, 
132  =  132. 

32.  Expressions  of  the  form  x^  +  sxi/  +  py^^  which  are  homo- 
geneous and  of  the  second  degree  with  reference  to  the  letters  x  and 
y,  may  be  factored  by  the  methods  employed  for  factoring  ^*  +  sx  +  p, 
by  separating  the  term  py^  into  two  factors,  each  containing  ?/,  the 
sum  of  which  is  the  coefficient,  s?/,  of  x  in  the  term  sxi/. 

Ex.  4.    Factor  x^  -\-l4xy  +  33 y^. 

The  term  33  y^  is  the  product  of  the  factors  II  y  and  3  y,  the  sum  of  which 
is  the  coefficient,  14 1/,  of  x  in  tlie  term  14  xy. 

Check. 

Hence  we  have  x^  +  14  x?j  +  Z3  y^  =  (x  i-  11  y)(x  +  3  y).      x  =  2,y  =  3. 

385  =  385. 

Exercise  XII.     11 

Obtain  factors  of  the  following  expressions,  checking  all  results 
numerically  : 

1.  JB*  +  ISic**  +  56.  12.  4:X^  +  20a;  +  9. 

2.  «*  +9a'+  18.  13.  4/  -2Sij+  13. 

3.  a"+  19«»+  90.  14.  9^^-  186  +  8. 

4.  b^+  5b*  +  6.  15.  25  c'  -  35  c  +  12. 

5.  ic^°  +  22a^  +  40.  16.  49c?'  +  Ud-  3. 

6.  x^  +  2x^  -  15.  17.  16/^'  +  40A  +  9. 

7.  ««-  18a;«+  72.  18.  8lP-45^-  14. 

8.  x"""  +  llic"  +  30.  19.  36i»2  +  24.x  — b. 

9.  x''"'  -  14af"  +  48.  20.  64/  -  80?/  -  11. 

10.  a;^'^  -  2 af  -  35.  21.  a"  ■\-lbah-\-  26  ^j^. 

11.  4a' +  16a  +  7.  22.  ar*  —  \2xij  -\-  21if. 


206  FIRST  COURSE  IN   ALGEBRA 

23.  P  —  19  ^c  +  34  c^.  27.  x"  ^2\xij  —  46 y«. 

24.  x'-2lxy  +  Z^y\  28.  a"  -  16ac  -  57  c\ 

25.  cr*-  12c</— 28fl?^.  29.  ar*  +  lOiKj^  — 56/. 

26.  J»^  4-  10  mn  —  39  «^  30.  h^-Vdhk  —  68  F. 

TYPE  IV.    Trinomials  of  the  Type  aai^  +  6a5  +  c 

33.  The  expression  ax^  +  6a;  +  c  is  called  the  general  expression 
of  the  second  degree  with  reference  to  a  single  variable,  jc,  because  by 
assigning  numerical  values  to  a,  b,  and  c,  the  expression  ax^  -\-  bx-\-  c 
may  be  made  to  represent  any  expression  of  the  second  degree  with 
reference  to  a  single  variable,  x. 

We  will  show  two  methods  for  factoring  expressions  of  the  type 

an^  +  bx  +  c. 

34.  First  Method.  A  method  for  obtaining  the  factors  of  an 
expression  of  the  type  ax"^  -\-  bx  +  c  may  be  obtained  as  follows  : 

If  the  expression  ax^  -\-  bx  ■\-  chQ  assumed  to  be  the  product  of 
two  binomial  factors  of  the  forms  mx  +  n  and  m'x  +  «',  the  product 
(mx  +  n){m'x  +  w'),  which  may  be  expressed  as 

'mm'x^  +  {mn'  +  m'n)x  +  nn'  must  represent  aa?  -\-  bx-\-  c. 

It  should  be  observed  that  the  term  {mn'  4-  m'n)  x  is  the  sum  of 
the  two  factors  mn'x  and  m'nx  of  the  product  mm'nn'x'^^  obtained 
by  multiplying  the  term  mm'x'^  by  the  term  nn'  which  is  free  from  x. 

It  follows  that  a  given  expression  of  the  type  ax"^  +  bx  +  c  can 
be  factored  by  inspection,  provided  that  the  product  acx^  can  be 
separated  into  two  factors  of  which  the  sum  is  bx. 

To  factor  a  trinomial  expression  of  the  type  ax^  +  bx  +  c,  multiply 
the  term  aoc^  by  the  term  c,  and  find  two  factors  of  the  product  ax'^c 
(each  containing  x)  of  which  the  sum  is  the  middle  term  bx. 

The  required  factors  of  the  trinomial  expression  may  then  be 
obtained  by  grouping  the  te7'ms  of  the  polynomial  expression  thus 
obtained. 

Ex.  1.     Factor  4a^+l5a+  14. 

The  product  obtained  by  multiplying  the  term  4a2  by  14  is  56  a^;  and 
56  a^  is  the  product  of  two  terms,  8  a  and  7  a,  of  which  the  sum  is  the 
middle  term,  15  a. 


INTEGRAL   FACTORS  207 

Hence  we  have, 

4a2-f  15a  + 14  =  4a2  +  8a  +  7a  + 14  Check,     a  =  2. 

=  4  a  (a  +  2)  +  7  (a  +  2)  16  +  30  +  14  =  15  •  4 
=  (4  a  +  7)  (a  +  2).  60  =  60. 

Ex.  2.    Factor  6  6«  -  13  6  -  8. 

Multiplying  6  b^  by  —  8,  we  obtain  —  48  6^  which  is  the  product  of  the 
factors  3  6  and  —  16  6  of  which  the  sum  is  the  middle  term,  —  13  6. 
Hence,  we  have, 

662-13  6-8  =  6  62  +  36-16  6-8  Check.     6  =  2. 

=  36(26  +  l)-8  (26+ 1)      24  -  26  -  8  =  -  2  •  5 
=  (3  6-8)(2  6  +  l).  -10  =  -10. 

35.  Second  or  Trial  Method.  Certainexpressionsof  the  type 
ax^  -\-  bx  +  c  may  be  factored  by  inspection  as  follows  : 

Separate  the  term  ax^  into  all  possible  pairs  of  factors^  each  factor 
containing  x ;  also  separate  c  into  all  possible  pairs  of  factors. 

For  the  first  terms  of  the  required  binomial  factors  of  ax^  -\-  bx  ■\-  c, 
use  the  factoids  of  one  of  the  pairs  of  factors  whose  product  is  ax^^  and 
for  the  second  terms  of  these  binomial  factors  use  the  factors  of  one 
of  the  pairs  of  factors  whose  product  is  c. 

The  different  pairs  of  factms  selected  for  first  and  second  terms 
of  tJie  required  binomial  factors  must  be  so  chosen  that,  when  the  bi- 
nomials are  arranged  as  multiplicand  and  multiplier^  the  sum  of 
the  cross  products  shall  be  the  middle  term  bx  of  the  expression 
ax^  -^^  bx  -\-  c. 

36.  For  convenience,  when  applying  this  method,  we  shall  sup- 
pose that  a  given  expression  of  the  type  ax^  ■\-  bx  +  c  has  been  so 
transformed  that  the  number  represented  by  a  is  positive.  The 
binomial  factors  may  then  be  so  chosen  that  their  first  terms  will 
be  positive.  Hence,  when  selecting  the  terms  for  these  binomial 
factors,  it  is  necessary  to  consider  only  the  signs  of  the  second 
terms. 

37.  If  the  third  term,  c,  of  an  expression  of  the  type  ax^  -\-  bx-{-  c 
be  positive,  the  signs  of  any  pair  of  factors  of  c  must  be  like.  Hence 
the  signs  of  the  second  terms  of  the  required  binomial  factors  are 
the  same,  and  must  be  like  the  sign  of  the  middle  term  of  the  given 
trinomial  which  is  the  sum  of  the  cross  products  containing  the  fac- 
tors of  c. 


208 


FIRST  COURSE  IN  ALGEBRA 


If  the  third  term,  c,  be  negative,  the  signs  of  any  pair  of  its  factors 
must  be  unlike.  Accordingly,  the  signs  of  the  second  terms  of  the 
required  binomial  factors  are  unlike. 

The  signs  of  these  second  terms  must  be  so  taken  that,  when  the 
required  binomial  factors  are  used  as  multiplicand  and  multiplier, 
the  sign  of  the  greater  cross  product  is  like  the  sign  of  the  middle 
term  of  the  given  trinomial. 

Ex.  3.    Factor  Gx^  +  13  a:  +  5. 

This  expression  is  of  the  type  ax^  +  fear  +  c,  6  corresponding  to  a,  13  to  6, 
and  5  to  c.  The  "  trial "  pairs  of  factors  of  6  x*  and  5  may  be  arranged  as 
follows: 


For  first  terms  of  the 
binomial  factors, 
3x. 
2x. 


we  have 


{'I 


For  second  terms  of  the 
binomial  factors, 

we  have    ■< 

The  first  terms  of  the  required  binomial  factors  must  be  terras  in  the 
same  column,  —  that  is,  either  6  x  and  x,  or  3  x  and  2  x,  respectively,  —  and 
the  second  terms  of  the  binomial  factors  must  be  5  and  1. 

We  may  now  arrange  the  different  pairs  of  factors,  6x  and  x,  3x  and  2x, 
as  first  terms  of  "  trial "  binomial  factors,  and  5  and  1  as  second  terms,  as 

follows : 

Selecting  one  pair  Interchanging  terms 
of  columns  of  second  column 

First  Binomial.        6x         +  5  6x         +  1 


Second  Binomial. 


First  Binomial. 


6  x2  +  1 1  X  +  5. 

Selecting  a  second 
pair  of  columns 
3x  +  5 


6  x2  +  31  X  +  5. 

Interchanging  terras 
of  second  column 
3x  +  1 


Second  Binomial.     2  x 


6x2  ^  13  X  +  5. 


2x         +  5 

6x2  +  17x  +  5. 


Examining  the  expressions  resulting  from  the   multiplications   of  the 

different  pairs  of  binomials  above,  we  find  that  in  the  first  expression  of 

the  second  row  we  have  obtained  the  product  Qx^  +  13x  +  5,  in  which 

the  middle  term  13xis  the  same  as  the  middle  term  of  the  given  expression. 

Hence,  we  may  write  Check.     Let  x  =  2. 

24  +  26  +  5=  U-5 
55  =  55. 


6x2+  I3x  +  5  =  (3x  +  5)(2x+  1). 


INTEGRAL  FACTORS 


209 


Ex.  4.    Factor  I2z^  +  8x-  15. 

This  expression  is  of  the  type  ax^  -\- bx  +  c,  12  corresponding  to  a,  8  to  6, 
and  —  15  to  c. 

The  sign  of  the  constant  term  —  15  is  minus,  and  accordingly  the  signs 
of  the  second  terms  of  the  binomial  factors  must  be  unlike,  one  positive 
and  the  other  negative. 

The  "trial"  pairs  of  factors  of  12 a;^  and  of—  15  may  be  arranged  as 
below,  the  double  signs  ±  and  =F  being  used  to  indicate  that  the  factors  of 
—  15  are  of  opposite  sign,  one  positive  and  the  other  negative : 


For  first  terms  of  the 
binomial  factors, 

,         (  12x,  Qx,  4x. 
we  have  ■{        '       ' 

(      a:,  2  a;,  3  a;. 


and  For  second  terms  of  the 

binomial  factors, 
±  15,  ±  5. 
=F    1,  =F3. 


we  have 


As  in  example  3,  we  may  construct  different  "  trial "  binomial  factors 
by  selecting  combinations  of  first  terms  and  second  terms  from  among  the 
diff'erent  columns  above. 

Since  in  all  cases  the  products  arising  from  the  multiplications  of  the 
first  terms  of  the  "trial"  binomials  as  arranged  below  are  12 a;^,  and  the 
products  of  the  second  terms  are  —  15,  we  have,  for  convenience,  shown 
only  the  cross-product  terms. 


With  a  given 
1st  column. 

1st  Binomial 
2nd  Binomial 
Cross  Product 

With  a  second 
Ist  column. 

1st  Binomial 

2nd  Binomial 

Cross  Product 

With  a  third 
Ist  column. 

1st  Binomial 
2nd  Binomial 
Cross  Product 


Selecting    Interchanging 
a  2nd  terms  of 

column.       2nd  column. 


New  2nd     Interchanging 
column.  terms  of 

2na  column. 


12a:  db  15 
XT  1 
±'Sx 


6x  ±  15 
2xT  1 
±24  a; 


4x±  15 

3a:  =F    I, 

±41  a: 


12a:  =F    1 

a:±  15 

±  179  a; 


6a:±  1 
2x±  15 
±88  a; 


4a:T    1 
3a:±  15 


±57a: 


12a:±  5 

a:=F3 

=F31a; 


6a:±5 
2a:T3 


4a:±  5 
3a:T3 
±  3  a: 


12a:T3 
a:±5 
±57  a: 


6a;=F3 
2a:±  5 
±24  a: 


4a;  T  3 
3a:  ±5 

±  11a; 


14 


210  FIRST  COURSE  IN  ALGEBRA 

The  middle  term,  8x,  of  the  given  expression  12x2 -f- 8ar  —  15  jg  ^he 
cross-product,  8  X,  which  arises  from  the  multiplication  of  the  trial  binomials 

6a:-5 
2a;  +  3 

+  8x. 
Hence  we  have,  Check.     Let  a:  =  2. 

48  +  16  -  15  =  7  .  7 
12x2  +  8x-15  =  (6a:-5)(2a:  +  3).  49^49. 

38.  It  will  not  usually  be  necessary  to  write  all  possible  com- 
binations of  trial  binomial  factors  as  in  Ex.  4,  above,  since  by 
inspection  we  may  perform  the  cross  multiplications  mentally,  and 
unsuitable  combinations  of  terms  may  be  rejected  at  once. 

When  constructing  the  different  binomial  factors,  we  should  be 
guided  by  the  following 

Principle :  If  no  monomial  factor  is  common  to  all  of  the  terms  qfa 
given  expression^  there  can  be  no  monomial  factor  common  to  the  terms 
of  any  polynomial  factor  of  the  given  expression. 

E.  g.  Since  no  monomial  factor  is  common  to  all  of  the  terms  of  the 
expression  12x2  +  8x  —  15  ^^^,^,  g^.  4  §  37),  it  follows  that  such  "  trial  " 
binomial  factors  as  12x  ±  li>,  12x  =F  3,  6x  ±  15,  Cx  =F  3,  3  x  ±  15  and 
3x  =F  3  must  be  rejected. 

Ex.  5.   Factor  8  x^  -  10  x  +  3. 

Observe  that,  since  the  sign  of  the  constant  term  3  is  +,  and  of  the  middle 
term  —  lOx  is  — ,  the  signs  of  the  second  terms  of  the  binomial  factors  must 
be  like,  and  both  — . 

The  factors  of  8  x^  and  of  3  may  be  arranged  for  first  terms  and  for  second 
terms  of  "  trial "  binomial  factors  as  follows : 

First  terms  Second  terms 

First  Binomial        >ar,    X>  ^  ^•iJ^t  —  3,     —X* 

Second  Binomial         ^J^^2x,  X"  —1,    —^ 

Cancelling  terms  which  we  find  cannot  be  used,  we  have  finally, 

8x2- 10x  +  3  =  (4x-3)(2x-l).  Check.     Let  x  =  2. 

32  -  20  +  3  =  5  •  3 
15=15. 

39.  Expressions  of  the  form  ax"^  +  hxy  +  cy"^-,  which  are  homo- 
geneous and  of  the  second  degree  with  reference  to  two  letters, 


INTEGRAL   FACTORS  211 

X  and  y,  may  be  factored  by  tbe  methods  employed  for  factoring 
expressions  of  the  type  ax^  +  hx  +  c. 

When  factoring  such  expressions  it  is  necessary  that  each  of  the 
factors  of  the  term  ax^  selected  as  the  first  terms  of  the  required 
binomial  factors  should  contain  x,  and  that  each  of  the  factors  of 
the  term  cif,  selected  as  second  terms,  should  contain  y. 

E.g.  The  expression  6  m^  +  25  m;i  +  lln^  may  be  expressed  as  the 
product  of  tbe  factors  3«t  +  11  ?i  and  2m  +  n. 

Exercise  XII.     12 

Factor  the  following  expressions,  checking  all  results  by  making 
numerical  substitutions  : 


1. 

3i«2+  4a;+  1. 

26. 

Slj-^-Ub-  39. 

2. 

2  x^  +  5  ic  +  2. 

27. 

7  a;'-  10a;  +  3. 

3. 

5.^-^  +  26a5+  5. 

28. 

30  6''-31c  +  8. 

4. 

4.x' +  lla;+  7. 

29. 

56  a;' +  65  a; -9. 

5. 

3i«-^  +  8aj  +  4. 

30. 

40?y'-|-  14?/  — 33. 

6. 

2a!' +  7a;  +  3. 

31. 

7 a;' 4-  13  +  20a;. 

7. 

lx^  +  33aj+20. 

32. 

28  a;' +  17  -48  a;. 

8. 

9a;'^+  34a; +  21. 

33. 

48a;' -23a;-  13. 

9. 

10a''+  19a;  +  9. 

34. 

18  a;' +  37  a;  +  19. 

10. 

6a;2  +  41a;  +  30. 

35. 

20  a'- 17a-3. 

11. 

7ar^+  13a;-2. 

36. 

8a'/>'+  103  a/; -13. 

12. 

8a'-5«-3. 

37. 

17  aV  — 69  aa;  +  4. 

13. 

3a'-5«-22. 

38. 

20?7i'+  ldmn  +  S7i\ 

14. 

6r*2~  19a  +  15. 

39. 

7  6'  +  4:lbc-Gc\ 

15. 

4a' +  8a -45. 

40. 

40  a;' -83  a;?/ +  42/. 

IG. 

18a;' +  9a; -4. 

41. 

dx^  —  ixy-  13/. 

17. 

2  a'-  15  a  — 8. 

42. 

9  a;' +  55  a;?/  —  56?/'. 

18. 

11  a' -21  a  +  10. 

43. 

6a'/>'+  7a6c-5c'. 

19. 

Ub^-b-  13. 

44. 

3(4ar*-21)-a;. 

20. 

13a'  — 2?  a  +  2. 

45. 

13  7w'-  (m+  14). 

21. 

14  w' 4-  73  7W  +  15. 

46. 

16 +  w  (l9?<;-34). 

22. 

14  a'  — 33  a  +  18. 

47. 

h(2'2h-  19)  —  15. 

23. 

54a'—  15a- 50. 

48. 

b(l  +  15  6)  -16. 

24. 

36  w'  +  27  m  +  2. 

49. 

2(3?/'-28;s')  +  41?/;r. 

25. 

30w'- 47  m- 5. 

50. 

7c  (29c?  +  14c)  +  14^. 

212  FIRST  COURSE  IN  ALGEBRA 

Type  V.  Biuomial  Sums  and  Differences  of  Like  Powers 

a»db6» 

40.   From  §§  55  -  58,  Chap.  VIII.,  we  have  the  following  principles: 

(i.)  The  stun  of  the  same  odd  powers  of  two  numhei^s  is  equal 
to  the  sum  of  tlie  two  given  numbers  multiplied  by  a  polynomial 
factor, 

(ii.)  The  flifference  of  the  same  odd  powers  of  two  numbers  is 
equal  to  the  difference  of  the  two  given  numbers,  multiplied  by  a 
polynomial  factor, 

(iii.)  The  difference  of  the  same  even  powers  of  two  numbers  is 
eqtuil  to  the  sum  of  the  two  given  numbers,  multiplied  by  a  poly- 
nomial factor,  or  is  equal  to  the  difference  of  the  two  given 
numbers,  multiplied  by  a  polynomial  factor. 

The  polynomial  factors  of  binomial  sums  or  differences  of  like 
powers  are  formed  according  to  the  Law  of  Quotients.  (See  Chap. 
VIII.  §  59.) 

Ex.  1.    Factor  m^  +  32. 

To  apply  the  principles  of  §  40,  it  is  necessary  that  the  terms  of  the 
binomial  be  expre.ssed  as  like  powers. 

Since  32  may  be  expressed  as  2^,  it  follows  that  m^  +  32  may  be  expressed 
as  m^  +  2^. 

The  binomial  m^  +  2^  is  the  sum  of  the  same  odd  powers,  of  m  and  2 
and  hence  by  Principle  (i.)  the  binomial  sum  w  +  2  must  be  one  of  the 
required  factors. 

Hence  we  have, 

wi5  _|_  26  =  [m  +  2]  [m*  -  m8  •  2  +  m2 .  22  -  m  •  28  +  2*] 
That  is,  w6  +  32  =  [m  +  2]  [m^  -  2  wi*  +  4  m^  -  8  m  +  16].     Check,    m  =  2. 

64  =  64. 

Ex.    2.    Factor  125  a:«-/. 

We  may  express  125x8  _  yZ  ^g  (Sx)'  —  y^,  which  is  the  difference  of  the 
same  odd  powers  of  5  x  and  y. 

It  follows  from  (ii.)  that  5  x  —  y  is  one  of  the  required  factors  of  this 
diflference. 

Hence  we  have, 

(bxy-y^  =  {bx-y']l{bxy^  +  (bx)y  +  y'^']    Check,   x  =  3,  ^  =  2. 

3375-8=13-259 
That  is,  125  x3  -  y8  B  [5  X  -  2/]  [25  x2  +  5  x?/  +  y^\  3367  =  3367. 


INTEGRAL  FACTORS  213 

Ex.  3.    Factor  ««  -  27  b^. 

The  binomial  a^  —  27  b^  may  be  expressed  as  (a^y  —  (3  6)3  which  is  the 
diflference  of  the  same  odd  powei-s  of  a^  and  3  b. 

By  Principle  (ii.),  one  of  the  factors  of  (a^y  —  (3  by  is  the  binomial 
difference  a^  —  3b. 

Hence  we  liave,  Check. 

(a2)8  _  (3  6)8  =  [a2  _  3  b][(a'^y  +  «2  (3  5)  _f.  (3  j)2-]  .       a  =  3,  6  =  2. 
That  is,  a«-27  68    =  [a^  -  3  b]la^ -\- 3 a%  +  9  b^].  513  =  513. 

Ex.  4.    Factor  ««  +  6«. 

The  binomial  «•  +  6*,  which  is  the  sum  of  the  same  even  powers  of 
a  and  6,  may  be  expressed  as  {a^y  +  (6^)*j  which  is  the  sum  of  the  same 
odd  powers  of  a^  and  b^. 

Considered  as  the  sum  of  the  same  odd  powers  of  a^  and  b^,  it  follows 
from  Principle  (i.)  that  one  of  the  factors  of  (a^y  +  (b^y  must  be  a^  +  b^. 

Hence  we  have, 

(a2)8  +  (b^y  =  [a2  +  62]  [(a^y  _  (a^)  (^2)  +  (h^y]         ciieck. 
That  is,     a«  4-  6«      =  [a^  +  6^]  [a*  -  a%^  +  b*].  a  =  2,  b  =  I. 

64  +  1  =  5(16-4+1) 
65  =  65. 

Exercise  XII.     13 
Factor  the  following  expressions,  checking  all  results  numerically  : 

1.  x^+  1.  17.  ««+  729  ^^ 

2.  a?  +  8.  18.  S2a^  —  2A3b\ 

3.  27r*'+  1.  19.  8x^  +  21  y\ 

4.  £c'-  1.  20.  32a^''+  1. 

5.  a^  +  1.  21.    1728  -27/. 

6.  32rt'+  1.  22.    125a;«-8/. 

7.  c«  —  64  (P.  23.   16  a*  —  625  b*. 

8.  a:«  -  64  y^  24.  81  x^  —  2561/. 

9.  81  a'  —  b\  25.  8  c«  +  343  ^8. 

10.  125  +  21  a\  26.  216^'  +  129  f. 

11.  a;^  — 2/1  27.  c?"  +  512 £c» 

12.  «i'  +  6^2.  28.  64a«+  1. 

13.  16a*-  1.  29.   128  a^-  1. 

14.  1000  a;«  —  ?/*.  30.  1024  a^  +  ^^ 

15.  343a»-^/«.  31.  3125  +  c'. 

16.  w''  +  128.  32.  1296  -  a\ 


214  FIRST  COURSE   IN  ALGEBRA 

33.  2401  -  x\  37.  o^  -  243/. 

34.  x^  —  216/.  38.  1000  7w»  -  z\ 

35.  64  a«  +  y.  39.  1728  a»  -  l}"". 

36.  /  -  512  2*.  40.  3125  c«  -  32  d^. 

Polynomials  of  at  Least  the  Third  Deg^ree  with  Reference 
to  Souie  Letter,  a; 

41.  Consider  the  following  product  of  binomials  : 
{x  —  a){x,  —  U){x  —  c)^^  —  {a-\-b-\-  c) x^  +  (ab  +  ac  +  bc)x~- abc. 

It  will  be  seen  that  the  first  term,  ^r*,  is  the  product  of  the  first 
terms  of  the  binomial  factors,  and  that  the  last  or  constant  term, 
--  abcy  is  the  product  of  the  second  terms,  —  a^  —  b  and  —  c,  of  the 
binomial  factors. 

It  may  be  seen  that  if  two  or  more  binomial  factors  of  the  form 
X  —  a  are  multiplied  together  to  form  a  product,  the  term  of  highest 
degree  with  reference  to  x  will  be  the  product  of  the  first  terms  of 
the  binomial  factors,  and  that  the  constant  term,  which  is  the  term 
free  from  .r,  will  be  the  product  of  the  second  terms  of  all  of  the 
binomial  factors. 

Hence,  if  binomial  factors  of  the  form  x  —  a  exist  for  any  given 
integral  polynomial  whose  highest  power  with  reference  to  x  has  the 
coefficient  unity,  we  may  discover  their  second  terms  among  tJis  factors 
of  the  term  that  is  free  from  the  letter  of  arrangement. 

We  may  accordingly  construct  different  ^^  triaV  binomial  factors, 
by  icriting  x  minus  each  of  these  factors  in  tmm.  Then  we  may 
apply  the  Remainder  Theorem  to  discover  which  of  these  ^' trial" 
factors,  if  any,  are  factors  of  the  given  expression. 

Ex.1.    Factor  x8- 6x2+ liar -6. 

Since  the  expression  is  of  the  third  degree  with  respect  to  ar,  we  shall  look 
for  three  binomial  factors  of  the  first  degree  with  respect  to  x,  having  the 
form  X  —  a. 

Since  the  term  free  from  x,  —  6,  is  negative,  the  signs  of  an  odd  number 
of  second  terms  of  the  binomial  factors  must  be  minus.  Neglecting  the 
sign,  the  integral  factors  of  6  are  1,2,  3,  and  6.  We  may  assume  as  "  trial " 
binomial  factors,  x— l.a:  —  2,  x  —  3,  and  x  —  6. 

It  may  happen  that,  of  the  final  set  of  three  binomial  factors  selected, 
two  binomials  may  be  sums  instead  of  differences. 


INTEGRAL  FACTORS  215 

Applying  the  Remainder  Theorem,  we  find  by  trial  that  a:  —  6  is  not  a 
factor  of  the  given  expression. 

It  will  be  convenient  to  use  the  method  of  synthetic  division  as  below : 

1     _6    +11      -    6     )+6 

+  6     +    0     +66 
1     +0    +11,     +60 

The  remainder  is  60.     Hence  the  division  is  not  exact. 
By  trial,  we  find  that  the  remaining  factors  are  contained  exactly  as 
divisors  in  the  given  expressio-       Hence,  we  may  write 

ars  _  6a;2  +  11 X  -  6  =  (x  -  1)  (ic  -  2)  (a;  -  3).         Check.     Let  x  =  4. 

64-96  +  44~6  =  3-2-l 
6  =  6. 

42.  After  having  found  one  factor  of  a  given  expression,  it  will 
often  be  possible  to  save  work  by  dividing  the  expression  by  this 
factor,  and  then  examining  the  quotient  for  the  remaining  factors 
of  the  given  expression. 

Ex.  2.    Factora:«  +  6a:2-a:-30. 

The  second  terms  of  such  binomial  factors  of  the  form  a;  —  a  as  may  exist 
will  be  found  among  the  integral  factors  of  —  30,  Since  the  expression  is 
of  the  third  degree  with  reference  to  x,  we  may  expect  to  find  three  bino- 
mial factors. 

The  integral  factors  of  30  are  1,  2,  3,  5,  6,  10,  15,  and  30.  Since  30  is 
negative,  at  least  an  odd  number  of  the  factors  of  30,  selected  for  second 
terms  of  the  required  binomial  factors,  must  have  minus  signs.  We  may 
assume  as  "  trial  "  binomial  factors  a;  —  1,  x  —  2,  a;  —  3,  a;  —  5,  a;  —  6,  a;  —  10, 
a;  -  15,  and  x  -  30. 

Applying  the  Remainder  Theorem,  we  find  that  a;  —  2  is  a  factor  of  the 
given  expression.  Since  x  —  2  is  a  factor  of  the  given  expression,  the  quo- 
tient obtained  by  dividing  the  expression  by  x  —  2  must  be  the  product  of 
the  remaining  factors. 

When  dividing  the  expression  x^  +  6  x^  —  x  —  30  by  x  —  2  it  will  be 
found  convenient  to  apply  the  method  of  synthetic  division,  as  shown 
below  : 

1+6-1      -30     )_2 

+  2     +  16      +30 
1-1-8+15,  0 

By  using  the  coefficients  1,  8  and  15  and  supplying  the  proper  powers 
of  X  we  may  immediately  construct  the  desired  quotient  x^  +  8x  +  15, 


216  FIRST  COURSE   IN   ALGEBRA 

Hence  we  may  write 

a:8  +  6x2  -  a;  -  30  =  (a;  -  2)(x^  +  8x+  15). 

The  trinomial  factor  x^  +  8  a;  +  15  is  the  product  of  the  binomial  factors 
X  +  3  and  x  +  5. 

Hence  we  have  finally,  Check.     Let  a;  =  3. 

a:8  +  6x=*-a;-30  =  (x-2)(x  +  3)(x  +  5).     27  +  54-3-30=1  -6 -8 

48  =  48. 

Exercise  XII.     14 

Obtain  factors  of  the  following  expressions,  checking  all  results 
numerically  : 

1.  x''^  6a;^+  lla;+  6.  11.  a*  +  4 a;^  -  28 a;  +  32. 

2.  a;»4-8ar'+  17aj4-  10.         12.  iB«  -  Ux^  +  51  a; -54. 
B.  a^-\-  8a;^+  19a;  +  12.         13.  x^  -  ISa;^  +  87a;  -  70. 

4.  x^+  na^+  34a;+  24.  14.  a;«  —  lOa;^—  17a;+  66. 

5.  a;«+  10a;^+  31  a;  +  30.  15.  a;«  -  11  a^  +  55a;- 39. 

6.  a;«+  8ar»+5a;-  14.  16.  a;»  -  3 ar^  -  34 a;  -  48. 

7.  a;»  +  3a;2-  13a;- 15.  17.  a;«  +  13a;' +  54a;  +  72. 

8.  x^'  +  x''-  22  a;  -  40.  18.  a;»  -  13  a;^  +  55  a;  -  75. 

9.  X*  —  2x^  —  29a;  —  42.  19.  a;*  -  3a;2  —  24a;  +  80. 
10.  a;^  —  27  a;  —  54.  20.  a;*  +  66  a;'  +  129  x  +  64. 

Determine  by  the  Remainder  Theorem  whether  or  not  a;  +  2  is  a 
factor  of  each  of  the  following  expressions  : 

21.  x^-{-  4a;'' +  2a; +  4.  27.  x^  —  10a;2+  27a;-  18. 

22.  a;*  —  a;'  -  5a;  +  2.  28.  a;^  +  lOa;'  +  29a;  +  20. 

23.  a;«  -  2a;2  +  4a;  -  6.  29.  x(x^  -  9)  -  2(3a;'  -  7). 

24.  a;«  +  2  a;2  -  29  a;  -  30.         30.  x(x^  -  10)  -  S{x^  -  8). 

25.  x^  +  Sx^+  3a;  +  2.  31.  a;«+  6a;2-a;-  18. 

26.  a;^  +  a;*''  —  10a;  +8.  32.  a;^  -  25 a;  4-  ar'  -  46. 

Suggestions  Relating  to  Methods 

43.  Although  no  rules  can  be  given  which  will  apply  in  all  cases, 
the  student  will  find  that  the  following  general  directions  will  serve 
to  systematize  the  work  when  factoring  a  given  expression  : 

(1)  Eemave  all  monomial  factors  which  are  common  to  all  of  the 
terms  of  a  given  e.rpression. 

(2)  Examine  the  resulting  expression  and  determine  whether  or 


INTEGRAL  FACTORS  217 

not  it  can  he  simplified  hy  applying  one  or  more  of  the  methods  shown 
in  this  chapter. 

Ex.  1 .   Factor   W  x"- +  22  xy  +  II  if. 

Dividing  all  of  the  terms  by  the  common  factor  11,  we  obtain  the  quotient 
x^  -\-2xy  -{■  y%  which  is  the  square  oi  x  +  y. 

Hence  we  have  Check.     Let  x  =  2,  y  =  3. 

Ux^  +  22xy-{-llf=n(x^-i.2xy-\-  f)     44  +  132  +  99  =  11  .  25 
=  ll(x  +  2^)2.  275  =  275. 

(3)  If  a  given  expression  is  arranged  according  to  descending  or 
ascending  powers  of  some  particular  letter ^  factors  will  frequently 
be  suggested  by  the  form  of  the  expression. 

Ex.  2.    Factor  x"^  {%i  —  a)  ■\- y'^  {a  —  x)  ■\-  o?  {x  -  y). 
Performing  the  multiplications,  and  rearranging  the  terms  according  to 
descending  powers  of  x,  we  obtain  the  expression 

a;2  (7/  -a)-x  (y^  -  a')  +  (ay^  -  ahj), 
in  which  the  binomial  factor  y  —  ah  common  to  the  different  groups  of  terms. 
Hence  we  have 

^^  (2/  -  »)  +  y^  {a-x)+  a'^  (x  -  y)  =  x\y  -a)-x  (y^  -  a^)  +  (ay^  -  ahj) 
Check.  =,x\y-a)-x{y  +  a){y-a)-{-ay{y-a) 

Let  a  =  1,  a:  =  3,  i/  =  2.  =  [x^  -x{y-\-a)-\-  ay]  [y  -  a] 

9  +  4(-2)+l=:2  =(^x-y){x-a){y-a). 

2  =  2. 

(4)  Occasionally  different  methods  of  factoring  may  be  applied  to 
different  groups  of  terms  of  a  given  expression.  It  may  happen  that 
these  groups^  after  being  separately  factored^  are  found  to  contain  one 
or  more  common  factors  which  in  turn  may  suggest  a  method  to  be 
applied  to  complete  the  factorization  desired. 

Ex.  3.     Factor  x^  •\- y^  -{■  ax  ■\-  ay  ■\-  bx  +  hy. 

In  the  given  expression,  the  binomial  sum  of  the  same  odd  powers  of  x 
and  y,  x^  +  y^,  may  be  expressed  as  {x  +  y){x^  —  ^U  +  y^),  and  the  remain- 
ing terms  ax  +  ay  -{■  bx  +  by  may  be  grouped  and  expressed  as  the  product 
{a  +  bj(x  +  y). 
Hence  we  have 
a;8  +  7/  +  ax  +  ay  +  bx  -{-by^  (x^  +  ?/)  +  [(ax  +  ay)  +  (hx  +  by)] 

Check.  =  (x  +  y)(x^  -xy  +  y'^)  +  (a  +  b)(x  +  y) 

Leta;  =  .3,  7/  =  2,  a  =  4,  &  =  5.       =ix^-xy  +  y'^  +  a  +  b)(x  +  y). 

80  =  16  •  5  ^      ^■ 

80  =  80. 


218  FIRST  COURSE  IN  ALGEBRA 

(5)  The  Remainder  Theorem  may  be  applied  when  a  given  expres- 
sion is  of  the  third  or  higher  degree  with  reference  to  some  letter  of 
arrangement. 

44.  When  for  any  reason  a  given  expression  appears  to  have 
more  than  one  distinct  set  of  prime  factors,  we  shall  find,  upon 
closer  examination,  that  factors  which  apparently  differ,  are  identi- 
cal. This  is  because  some  of  the  factors  which  were  supposed  at 
first  to  be  prime  still  admit  of  further  reduction,  or  else  differ  only 
in  sign. 

Ex.  4.    Factor  a:*  —  y*. 

Using  different  methods,  we  obtain 

Method   I.    (See  §  22)     x*  _  y*  =  (a;2  ^  yi^(^x  ^  y)(x  -  y). 

Method  II.   (See  §  40)     x^-t^\  "^'^'^  ^^  +  'If,  7  ^  |  ""'l  7  '2' 

Check.     Let  x  =  3  and  y  =  2. 
Method  I.     81 -16=  13-5 -1 
65  =  65. 

Method  IL   8l-ie=\'^^''^'ll^ 
(     or     1  •  65. 

65  =  65. 

Neither  of  the  polynomial  factors  obtained  by  Method  II  is  prime. 
Considering  the  first  polynomial  factor,  x^  —  x^y  +  xy^  —  y^,  we  find  that 

a*-xhf-{-xy^-y^={x^-yf^)-  {xh)  -  xy'') 

=  (a^  -  y){^^  +  a:y  +  2/2)  -  xy{x  -  y) 
=  [(a:2  +  xy  +  7/)  -  xy]lx  -  7/] 
=  (x^  +  y^Xx-y). 

Accordingly,  the  first  set  of  factors  obtained  by  Method  II  reduces  to 

(X  +  y)(x^  +  y^Xx  -  y). 
The  factors  in  this  set  are  identical  with  those  obtained  by  Method  I. 
In  a  similar  way  we  may  show  that  the  second  set  of  factors, 

(x  -  y)(x^  +  x^y  +  X7/2  +  y^), 
obtained  by  Method  II,  may  also  be  reduced  to  the  set  of  factors 
(x-yXx^  +  y^)ix  +  y). 

Ex.  5.    Factor  1ax-3a^-2x^. 

We  may  place  the  expression  in  minus  parentheses  and  arrange  the  terms 
according  to  descending  powers  of  a,  and  factor  as  follows ; 


INTEGRAL   FACTORS  219 

7ax-3a^-2x^  =  ~  (3a^  -7ax  -\-2x^)        Check.     Let  a  =  3,  x  =  2. 

=  — (Sa  —  x)(a —  2x).  7  =  7. 

If,  however,  the  expression  be  placed  in  minus  parentheses  and  the  terms 
be  arranged  according  to  descending  powers  of  a:,  we  have 

7 ax  -  3a^  -  2x^  =  -  (2x'^  -  7 ax  +  Sa^)       Check.     Let  a:  =  2,  a  =  3. 

=  -(2x-a)(x -3a).  7  =  7. 

We  shall  find  that  the  fnctors  first  obtained,  —  (3a  —  x)(a  —  2x),  difi'er 
from  tlie  factors  —  (^2 x  —  a){x  —  3  a)  only  in  the  signs  of  the  terms  of  the 
binomials. 

We  may  show  that  the  factors  of  the  second  set  are  identical  with  those 
of  the  first  set,  as  follows  : 

-(2x-  aXx  -  3a)  =  -  (2x  -  a)(x  -  3a)(-  1)(-  1) 
=  _  (2  a:  -  a)(-  1  )(x-  -  3  a)(-  1) 
=  _  (_  2 a:  +  «)(-  a;  +  3  a) 
=  —  (a  —  2  a;)(3  a  —  x). 

45.  Certain  expressions  may  be  factored  by  applying  any  one  of 
several  different  methods. 

Ex.  6.    Factor  16  a*  -  41  aV  +  25  c*. 

First  Method.  On  examination,  we  find  that  two  of  the  terms  16  a*  and 
25  c*  are  squares.  A  trinomial  square  having  these  same  two  terms  would 
have  as  a  middle  term  twice  the  product  of  the  square  roots  of  these  terms, 
—  that  is,  40  oV.  The  sign'  of  this  term  will  be  plus  or  minus  according 
as  the  trinomial  is  the  square  of  a  sura  or  the  square  of  a  difference. 

Solution  I.  Assuming  first  that  the  middle  term  is  +  40  a^c%  we  shall 
obtain  the  factors  as  follows  : 

The  difference  between  40  a^c^  and  —  41  aV-^  is  81  aV. 
Accordingly  we  have, 
16  a*  -  41  aV  +  25  c*  =  16  a*  -  41  «%«  +  25  c*  +  81  aV  _  81  a^c^ 
=  16  a-*  +  40  aV  +  25  c*  -  81  a%2 
=  (4a2  +  5c2)2-  (9acy 
=  [4  a2  +  5  c2  +  9 a6-][4 a^  +  5  c^  -  9  ac'] 
=  (4a  +  5c)(a  +  c)(4a  -  5c)(a  -  c). 
Solution  II. 

Assuming  secondly  that  the  middle  term  is  — 40  a^c^,  we  obtain  the 
same  result,  as  follows ; 

16  a*  -  41  oV  +  25  c*  =  16  a*  -  41  a'^c^  +  25  c*  +  a^c^  -  a^c^ 
=  16  a*  -  40  aV  +  25  c*  -  aV 
=  (4  a2  _  5  c2)2  _  (^ac)^ 
=  [4a2  _  5 c2  +  ac]  [4a2  -  5 c^  -  ac"] 
=  (4  a  +  5  c)(a  -  c)(4  a  -  5  c)(a  +  c). 


220  FIRST  COURSE   IN  ALGEBRA 

Second  Method.  We  may  apply  the  methods  of  §§  33-39  to  the 
expression  iis  given,  and  thus  obtain  the  same  factors,  as  follows  : 

16 a*  -  41  aV  +  25  c*  =  (16  a^  -  25  c^)Ca^  -  c^) 

=  (4a  +  5c)(4a-5c)(a  +  c)(a  -  c). 

Check.     Let  a  =  3,  c  =  2. 
1296  -  1476  +  400  =  22  •  2  •  5  .  1 
220  =  220. 

46.  Any  homogeneous  function  of  two  letters  may  be  factored, 
provided  that  it  is  possible  to  factor  the  non-homogeneous  expres- 
sion resulting  from  giving  the  value  unity  to  one  of  the  letters. 

Ex.  7.    Factor  x^  -  0  xhj  +  26  xy^  -  24  y^. 

By  assigning  the  value  1  to  i/,  the  expression  x^  —  0  x^y  +  26  xij^  —  24  y*, 
which  is  homogeneous  with  reference  to  x  and  i/,  reduces  to  the  non- 
homogeneous  expression  z*  —  9  x-^  -f  26  x  —  24. 

The  expression  x*  —  9  a;"^  +  26  x  —  24  may  be  factored  as  follows  : 

x«  -  9x2  +  26x  -  24  =  (x  -  2)(x  -  3)(x  -  4). 

From  this  identity  we  may  obtain  the  factors  of  the  given  expression  by 
introducing  such  powei-s  of  y  as  are  necessary  to  make  the  members  homo- 
geneous expressions  with  reference  to  ar  and  y. 
Hence  we  have. 

Check 
a;8_9a;2y  4.26XJ/2-  24?/8=  (x  -  2ij)(x -3y)(x-4ij).     Letx=:3,y  =  2. 

27  -  162  +  312  -  192  =  (-  1)(-  3)(-  5) 
-  15  =  -  15. 

Exercise  XII.     15     Miscellaneous 
Obtain  factors  of  the  following,  checking  all  results  numerically : 

1.  a«  -f  a^  -I-  a.  11.  49F-  36/. 

2.  xif  +  Ax-Sij-  12.  12.  x^^-f"", 

3.  a^  +  22a  +  121.  13.  ar^  -  22£c  +  105. 

4.  ar»  -  144.  14.  si?  +  a^  —  x  —  I. 

5.  a^  -t-  3ic  +  2.  15.  ii(^  +  ex +  2  dx+  2  cd. 

6.  2ar^+ lla;+ 12.  16.  (a-by-U  (a-b) -12, 

7.  3^2-12.  17.  a«-343. 

8.  3^  4-  33^  +  72.  18.  2a*  +  54. 

9.  km  +  2lm  —  sk  —  2  Is.  19.  1  a*  —  1  a. 

10.  ar^4-  5a;-6.  20.  5m -f  6m^+  1. 


INTEGRAL   FACTORS 


221 


21.  m^  +  m-2. 

22.  2a^  +  2Sa^+  66  a. 

23.  3a' +  30a  +  27. 

24.  13  x"  +  25  a;  — 2. 

25.  2a;2+  12ic+  18. 

26.  2«2  +  3a  +  1. 

27.  6  a'  — 3  a  — 3. 

28.  169  c' -9  c?'. 

29.  56  —  15  a;  4-  x\ 

30.  15a;' +  2xij-24.y\ 

31.  75a'-3^''. 

32.  5a;' +  20a;  +  20. 

33.  15  a' +  41a  +  14. 

34.  a  +  a\ 

35.  a;«+  19a;«  +  88. 

36.  a'6'+  30a6+  104. 

37.  a'x'+  3abx  +  2b\ 

38.  14  — 21m—  14  m'. 

39.  128  m*  -  18  n". 

40.  c*-5c'c?'  +  4g?^ 

41.  27c'4-  18c+  3. 

42.  ^y^z^-^xyz-  12. 

43.  8  a'h^c^  -  18  c^ 

44.  a;*- 21a;'+  80. 

45.  25a^<'-26a'+  1. 

46.  144  a;'- 625/. 

47.  16a^-41«'c'  +  25  c^ 

48.  64  w"  +  2. 

49.  8  a9  +  729. 

50.  6'+  288  -34^>. 

51.  a;«  +  25a;^  +  24. 

52.  50  — 20a;  +  2a;'. 

53.  147  a'-  75. 

54.  x^  —  38  a;*  +  105. 

55.  8  -  9  a;*  +  «'. 

56.  64  — a*. 

57.  6a'+  150  — 60  a. 


58.  3a;'+  36a^^+  108/. 

59.  4  a;'  — 28  a;?/ +  49/ 

60.  80  a'- 20a'^>'. 

61.  64r'  +  80r5  +  25  s'. 

62.  a;»—  7  a;' +  14  a;  — 8. 

63.  (a  +  hf  -  1. 

64.  9a'+  24a6+  16  Z>'. 

65.  (a  +  ^>)'  +  8  (a  +  ^)  +  15. 

66.  (a  +  ^)'  4-  5  (a  +  />)  +  6. 

67.  3c'—  14c?/ +  8/ 

68.  a""  +  2a'"^'"  +  6'"*. 

69.  30  a'—  154  a  +  20. 

70.  a;""  -  /. 

71.  (c  +  ^'  +  12  (c  +  of)  +  20. 

72.  a'— 2a^>  +  ^'-lla+ 11^^-12. 

73.  4  a;*-  13  a;'/ +  9/. 

74.  a«-50a»+  49. 

75.  a;«+  9  a;'  +  26a;+  24. 

76.  a«  -  9  a%  +  23  ah"-  -  15  6^ 

77.  (a  +  6)'  -  2  (a  +  ^)  +  1. 

78.  c?*+  11  (2  6^'+  11). 

79.  18^'- 31M+  6 A:'. 

80.  16  a' -(2  6  +  3  c)'. 

81.  a«-  13  a  +  12. 

82.  12  a'—  \2h{:la  —  h). 

83.  (c  +  ^'  +  10  (c  +  cO  +  21. 

84.  2a(a  +  6)  +  18. 

85.  a;'?/'^;'  —  8  xyzw  —  20  w'. 

86.  4m  (m  +  3)+  9. 

87.  2a  (4a  —  19)+  35. 

88.  (2a;  +  3?/)'-(3a  +  h)\ 

89.  x^  —  y^  —  2ijz  —  z\ 

90.  49  m*  —  65  m'7^'  +  16  n\ 

91.  a'—  7  (2a—  7). 

92.  a^  +  8  a. 

93.  /'''-68/'-  140. 

94.  A  X  {x  -\-  y)  +  ?/'. 


222  FIRST  COURSE  IN  ALGEBRA 


95. 

3^-2  +  4(2r+  1).                 98.  c^  +  (P  J^  2cd~a^~h^-2ab. 

96. 

0,8  _  3  a-/  +  2  /.                  99.    c^  -2cd+  d'  -  2(c  -  d)e  +  e\ 

97. 

a"  +  49^'^  -  I  +  14«^>.     100.   x'^-  (c-\-2d)x-\-  2cd. 

101.   a*_a«_  7^2+  a  +  6. 

102.   m^+n''+2mn-  a"  -  h'' -  2 ah. 

103.    1  -ir^cd-o'-dK 

104.    h""  +  2hm  +  711"  +  2hy  +  2my  +  y\ 

105.    16  a;*- 81 7/^ 

106.   9  +  49  a^- 58  a;*. 

107.   aj'-9a;. 

108.   «2^«-  16/>V. 

109.   aj*  +  //  +  a  +  2^. 

110.  x'-y'-{x  +  y)(:x-y). 

111.  a'^^H-  3a/>'— 3rt»-<^». 

112.  a"  +  ^''  -  c*  -  w^  -  2a/;  +  2  7»c. 

113.  4a^»r-  32af*y  +  64/". 

114.  7?  +  2xy  +  y''  +  'tixz  ^  Hyz  +  l^zy 

115.  4a^-2562+  2a  +  5  6. 

116.  xYz^-Haxyz-'20a\ 

117.  «*  +  9rt^/>*+  18/;*. 

118.  /'"-  168/* -340. 

119.  a%^  -  c^b^  -  a^d^  +  c^d^. 

120.  a^^+46^^-5a«/>«. 

121.  aft  (ar»+  1)  +«(«'* +  ft2). 

122.  1  +  6- 56  6^. 

123.  1  —  17a;2+  16  a;*. 

124.  4  — 52a;'^+  169  a;". 

125.  ahc^  +  3  ahc''  —  abc  —  S  ah, 

126.  4  (a"  +  c^jCa'^  -  c^)  +  3  ft'  (4  a^  -  4  c^), 

127.  a«  -  ft«  -  a(a2  -  ft^)  +  ft(a  -  bf. 

128.  9a^+  71a'  — 8. 

129.  (x  +  y){x  +  y-\-  7)  +  10. 

130.  a*-ft2(lla2-ft2). 

131.  w^2«_  19  ^«  +  34^. 

132.  a^^^^-  18a^ft^-  144  ft^**. 

133.  hcx^  +  (ac  +  ft^  a;  +  a^. 

134.  T?"^  +  (a  +  b)af"  +  ab. 


INTEGRAL   FACTORS  223 

135.  25a'  +  81b^+  Ua'^b*, 

136.  a^  (a  +  1)  +  ^^^  +  1)  +  2  ab. 

137.  x^  lx+  1)  —  2xij  —  f  {y  -  1). 

138.  a""  (a  +  3b)  +  P  (b  +  Sa). 

139.  x^+  2ax  +  a^  —  x-a. 

140.  8  a«  —  8. 

141.  16  «^'-  16. 

142.  12a^^+  12. 

143.  6«'^  +  5ax  +  x^. 

144.  14  a^-  109  «^>- 24^2. 

145.  7x^  +  ^1x1/+  30/. 

146.  a2-6P-^>(2a-6) +  c(2  6?-c). 

147.  c^  +  d^-  e^  -r  +  2  (^/-  cfl?). 

148.  a!*-2a;8+2ic—  1. 

149.  w'x  +  6V  +  b^x  ■\-aSj  +  2  (abx  +  a%). 

150.  (a  +  6)2  +  (6  +  dy  -(c  +  dy  -  (c  +  «)2. 

Application  of  the  Principles  of  Factoring  to  the 
Solution  of  Equations 

47.  To  solve  an  equation  containing  one  unknown  is  to  find  such 
a  value,  or  values,  for  the  unknown  as  will,  when  substituted  for  the 
unknown,  make  the  two  members  of  the  equation  identical. 

Ex.1.    Solvea;2+15  =  8a;.  (1) 

By  transposing  the  term  8  a;  to  the  first  member  we  have, 
a;2_8a;+15  =  0. 
Factoring,  (x  -  5)  (a:  -  3)  =  0.  (2) 

Sy  §§  27,  25,  Chap.  X.,  the  derived  equation  (2)  is  equivalent  to  the 
original  equation  (1). 

If  either  of  the  factors  a:  —  5  or  x  —  3  becomes  zero  for  any  particular 
value  of  a:,  the  other  remaining  finite,  the  product  of  the  two  factors  will 
become  zero.  Hence  for  such  a  value  of  x  the  first  member  of  the  equa- 
tion will  take  the  same  value  as  tlie  second,  that  is,  it  will  become  zero. 

If  X  be  given  either  of  the  values  5  or  3,  one  of  the  factors  will  become 
zero,  and  the  other  a  finite  number. 

Accordingly  these  values  are  solutions  of  equation  (2). 

It  may  be  seen  that,  by  placing  the  factor  x  —  b  of  the  first  member  of 
equation  (2)  equal  to  zero  and  solving  the  equation  thus  formed,  we  shall 
obtain  x  =  5,  which  is  one  of  the  solutions  of  equation  (2). 


224  FIRST  COURSE  IN  ALGEBRA 

It  appears  that  the  values  for  the  unknown  may  he  found  hy  solv- 
ing the  separate  equations  formed  by  ivriting  equal  to  zero  each  of  the 
factors  of  the  first  member  of  the  equation  obtained  by  transposing  all 
of  the  terms  of  the  given  equation  to  the  first  member. 

Accordingly,  from  (2)  we  may  write 

(x  -  5)  =  0,     also     {x  -  3)  =  0, 
Hence,  x  =  5,  x  =  3. 

By  substitution,  these  values  are  found  to  satisfy  the  original  equation. 
Ex.  2.    Solve  x«  +  x2  =  12  X.  (1) 

Transposing  and  factoring,  we  obtain  the  equivalent  equation 

x(x  +  4)(x  -  3)  =  0.  (2) 

Since  this  equation  is  satisfied  by  any  value  of  x  which  makes  any  one  of 
the  factors  of  the  first  member  zero  (the  other  factors  remaining  finite),  we 
may  place  each  of  the  factors  of  (1)  equal  to  zero,  and  solve  the  resulting 
equations. 

x  =  0,  x  +  4  =  0,  x-3  =  0. 

Therefore,         x  =  0,  x  =  —  4,         and     x  =  3. 

These  values  are  all  solutions  of  the  given  equation. 


Substituting  0 

for  X  in  (1), 

0  =  0. 


Sub.  —4  for  X  in  (1), 
(_  4)8 +  (-4)2  =12  (-4) 

-  48  =  -  48. 


Sub.  3  for  X  in  (1), 
38  +  .32  =  12  •  3 
36  =  36. 


48.    These  examples  illustrate  the  following 

Principle  of  Equivalence :  If  the  terms  of  an  integral  equation 
he  all  transposed  to  one  member,  and  if  this  member  be  factored  and 
the  sejMrate  factors  be  placed  equal  to  zero,  tlie  set  of  equations  thus 
obtained  will  be  equivalent  to  the  original  equation. 

That  is,  in  particular,  the  equation 

{x  —  a){x  —  b){x  -  c)  =  0,  (1) 

is  equivalent  to  the  following  set  of  equations  : 

X  —  a  =  0  X  —  b  =  0j         and  x  —  c  =  0.  (2) 

Solving  these  equations  separately  we  obtain  the  following  solu- 
tions of  the  given  equation  : 

X  =:  a,  x^  b,         and  x==  c. 


INTEGRAL  FACTORS  225 

The  equivalence  may  be  established  as  follows  : 

It  may  be  seen  that  when  x  is  given  any  value  which  reduces  one 
of  the  factors  of  the  first  member  of  equation  (1)  to  zero,  the  same 
value  will  also  reduce  the  first  member  of  one  of  the  equations  of 
set  (2)  to  zero.  Accordingly  such  a  value  of  a;  satisfies  equation  (1), 
and  also  one  of  the  equations  of  set  (2) . 

Since  the  factors  of  the  first  member  of  equation  (1)  are  the  first 
members  of  the  equations  of  set  (2),  it  follows  that  every  solution 
of  equation  (1)  must  also  be  a  solution  of  one  of  the  equations  of 
set  (2). 

Furthermore,  any  value  which,  when  substituted  for  a?,  satisfies 
one  of  the  equations  of  set  (2),  must  reduce  the  first  member  of  one 
of  the  equations  of  set  (2)  to  zero.  Accordingly  such  a  value  of  x 
will  reduce  one  of  the  factors  of  the  first  member  of  equation  (1)  to 
zero,  and  hence  will  satisfy  equation  (1). 

Hence  solutions  are  neither  gained  nor  lost  in  passing  from  the 
single  equation  (1)  to  the  set  of  separate  equations  (2)  ;  that  is, 
equation  (1)  is  equivalent  to  the  set  of  equations  (2). 

49.  If^  after  having  transposed  all  of  the  terms  of  an  equation  to 
one  member,  it  is  possible  to  separate  the  resulting  member  into  factors^ 
we  may  completely  solve  the  original  equation,  provided  that  these 
factors  are  of  such  forms  that  we  are  able  to  solve  the  equations 
formed  by  equating  them  separately  to  zero. 

50.  According  as  the  unknown  quantity  appears  in  an  integral 
rational  equation  to  the  first,  second,  third,  or  fourth  powers,  the 
equation  is  said  to  be  linear,  quadratic,  cubic,  or  biquadratic. 

Ex.  3.    Solve  the  quadratic  equation  2  a:^  =  a:  +  6. 
Transposing  the  terms  to  the  first  member,  we  have, 
2a:2-a;-6  =  0. 
Factoring,  (2  a:  +  3)  (a;  -  2)  =  0. 

This  single  quadratic  equation  is  equivalent  to  the  following  set  of  two 
linear  equations: 

2  a;  +  3  =  0,     and     a:  —  2  =  0. 

Solving  these  equations  separately,  we  obtain  the  following  values  which 
are  the  required  solutions  of  the  original  equation : 

ar  =  —  |,       and      a;  =  2. 
16 


226  FIRST  COURSE  IN   ALGEBRA 

Verifying  these  solutions  by  substituting  in  the  original  equation,  we 
have: 

Substituting  —  |  for  x,  Substituting  2  for  z, 

2  (-  1)2  =  (_  I)  +  6  2  (2)2  =2  +  6 

I  =  |.  8  =  8. 

Exercise  XII.     16 

Solve  the  following  equations,  regarding  the  letters  appearing  in 
them  as  unknowns,  and  verify  all  solutions  by  substituting  for  the 
letters  in  the  original  equations  the  particular  values  found  : 


1.     X'-4:X+S   =  0. 

27. 

^^  =  -10^. 

2.  i/  —  5y+  ^  =  0. 

28. 

P  -  4  =  0. 

3.  a'^- 6a- 7  =  0. 

29. 

a2  _  9  ^  0. 

4.  b^-7b+  10  =  0. 

30. 

b^-  16  =  0. 

5.  c^-8c+  12  =  0. 

31. 

v^  =  25. 

6.  g^'—^g^  18  =  0. 

32. 

m^  =  36. 

7.  A^  4-  5^+  6  =  0. 

33. 

71^  =  49. 

8.  P+  6^+  8  =  0. 

34. 

(P=d, 

9.   T/i"  +  4  7?2  —  5  =  0. 

35. 

k'-k  =  Q. 

10.  n^+  In  — S  =  0. 

36. 

h'-h  =  12. 

11.  z^+  nz+  30  =  0. 

37. 

7-2  +  r  =  20. 

12.  2t'2+  lli«;4-  24  =  0. 

38. 

6-2  —  s  =  42. 

13.  c^-\-  llc+  10  =  0. 

39. 

x"  +  x  =  56. 

14.  g""-  10  g  +  25  =  0. 

40. 

t^-  t  =  72. 

15.  r^  +  14  r  +  49  =  0. 

41. 

2/^ +22/ =15. 

16.  s^- 18s4-  81=0. 

42. 

;^2+  3;S  =  28. 

17.  x'+  12a; +36  =  0. 

43. 

^2  +  4  «^  =  45. 

18.  f—  16^ +  -64  =  0. 

44. 

a^-\-  Qa  =  U. 

19.  z^-2z  =  0. 

45. 

b-2—  !jb  =  50. 

20.  w^  —  3w  =  0. 

46. 

c2  _  7  c  =  18. 

21.  c2  +  4c  =  0. 

47. 

^2_8^  =  48.    • 

22.  ^  =  6  d. 

48. 

A2  +  12  =  7  A. 

23.  m^=lm. 

49. 

P  +  40  =  13^. 

24.  5n  =  n^ 

50. 

m^-\-  32  =  12  m. 

25.  Sp=2y^' 

51. 

n^+  13  =  14;^. 

2Q.  q'  =  -^q. 

52. 

r*—  14  =  5r. 

INTEGRAL   FACTORS 


227 


53.  5^-21=45. 

54.  t''—2Q>  =  lit. 

55.  -y'  — 35  =  2v. 

56.  ^>2_38==  17  ^,. 

57.  F-48  =  13^. 

58.  A^  -  33  =  -  8  A. 

59.  m^  —  34  =  —  15  7W. 

60.  7-=*  —  27  =  —  6  r. 

61.  22^-7^  +  3  =  0. 

62.  3g^-lg  +  2  =  0. 

63.  13  w^  -  14  «  4-  1  =  0. 

64.  5<^  — 21fi?+  4  =  0. 

65.  7«<;^-  15i^;  +  2  =  0. 

66.  liar*- 34a;  +  3  =  0. 

67.  6f—  11/ +  2  =  0. 

68.  Iz^—  10z+  3  =  0. 

69.  2w^  —  5w+S  =  0. 

70.  3a^4-  4a+  1  =0. 

71.  56'  +  6^+  1  =0. 

72.  7  c'  +  22  c  +  3  =  0. 

73.  llG?*'  +  23c?+  2  =  0. 

74.  55r2+7^+2  =  0. 

75.  6F+  rjk+  1  =0. 

76.  10w'+  7w+  1  =0. 


77.  15  72^  +  8;i+  1  =0. 

78.  20r'+  12r+  1  =  0. 

79.  20s2+  95+1=0. 

80.  2U2+  10^+  1  =0. 

81.  18  a;'—  11a;  +  1  =0. 

82.  26/-  15^^+  1  =  0. 

83.  305^2-  13;2;+  1=0. 

84.  ««7«  =  64w. 

85.  a«  =  81a. 

86.  6«  =  100  b, 

87.  c»  =  cK 

88.  c'  -  6  c'  -  7  c  =  0. 

89.  (P-ld^-Sd  =  0. 

90.  h^+  Sh^-  I0h  =  0. 

91.  F  +  4F—  12A:  =  0. 

92.  m^  +  Qm^—lm  =  0, 

93.  w«  -  8  w'  -  9  /i  =  0. 

94.  /+  lOj?'- llp  =  0. 

95.  ^«  +  4^2-21^  =  0. 

96.  ?•*—  llr'  +  28  =  0. 

97.  a;*  -  12  ar'  +  35  =  0. 

98.  i/-  lSf+  36  =  0. 

99.  z^+  9;.'+  14  =  0. 
100.  w^+  10  2^'+  16  =  0. 


228  FIRST  COURSE   IN   ALGEBRA 


CHAPTER  XIII 

HIGHEST  COMMON  FACTOR 

1.  In  this  chapter  we  shall  undertake  to  find  whether  or  not  a 
rational  integral  expression  can  be  found  by  which  two  or  more  given 
expressions  which  are  rational  and  integral  with  reference  to  certain 
letters,  «,  6,  c,  x,  y,  z^  etc.,  can  be  divided. 

2.  Two  given  integral  expressions  are  said  to  be  prime  to  each 
other  if  there  exists  no  expression  which  is  integral  with  reference 
to  the  letters  involved  by  which  the  given  expressions  may  be 
divided  without  remainder. 

E.  g.  x^  -f-  3  a:  +  5  and  x'^  +  2  a;  +  3  are  prime  to  each  other  because  no 
monomial  or  polynomial  divisor  can  be  found  by  which  both  of  these 
expressions  can  be  divided  without  remainder. 

3.  A  common  factor  of  two  or  more  integral  algebraic  expres- 
sions is  an  integral  expression  by  which  each  of  them  can  be  divided 
without  remainder. 

4.  The  highest  common  factor  (H.  C.  F.)  of  two  or  more 
integral  algebraic  expressions  is  the  product  consisting  of  the  entire 
group  of  factors,  numerical  and  literal,  by  which  each  of  the  given 
expressions  can  be  divided  without  remainder. 

5.  From  this  definition  it  appears  that,  since  the  highest  common 
factor  is  the  entire  common  factor,  it  must  be  of  the  highest  possible 
degree  with  reference  to  any  particular  letter. 

When  the  given  expressions  are  monomials,  the  highest  common 
factor  must  contain  the  numerical  greatest  common  divisor  (G.  C  D.) 
of  the  numerical  coefficients. 

When  the  given  expressions  are  polynomials,  the  highest  com- 
mon factor  may  be  the  product  of  a  monomial  and  a  polynomial  fac- 
tor. In  that  case  the  monomial  factor  is  the  highest  common  factor 
of  all  of  the  terms  of  the  given  expressions,  if  they  have  common 
factors;  the    polynomial   factor  is   the  polynomial   expression  of 


HIGHEST   COMMON   FACTOR  229 

highest  degree  with  reference  to  some  particular  letter  by  which 
both  of  the  given  expressions  can  be  divided  without  remainder. 

6.  When  the  given  expressions  are  monomials,  the  degree  of  the 
highest  common  factor  is  usually  reckoned  by  taking  into  account 
all  of  the  letters  entering  into  it. 

Highest  Common  Factor 
By  Factoring 
Monomial  Expressions 

Ex.  1.    Find  the  highest  common  factor  of  a^xhjh  and  a^x^yho. 

A  divisor  may  be  constructed  containing  the  letters  a,  x,  and  y,  which  are 
common  to  the  given  expressions. 

Observe  that  a  is  found  in  each  of  the  expressions  to  at  least  the  fourth 
power,  X  to  at  least  the  second  power,  and  y  to  the  third  power. 

As  there  can  be  no  common  factor  of  higher  degree  with  reference  to  any 
of  the  letters,  the  entire  common  factor  is  a^x'^y^. 

The  degree  of  this  highest  common  ftictor  may  be  reckoned  in  terms  of 
any  particular  letter,  but  it  is  usual  in  such  cases  to  reckon  it  in  terms  of 
all  of  the  letters. 

Accordingly  the  highest  common  factor  a*x^y^  is  considered  as  being  of 
the  ninth  degree. 

Ex.  2.    Find  the  H.  C.  F.  of  a%\'^,  a%k^  and  a^^c. 

The  highest  common  factor  is  a%^c,  for  each  of  the  letters  a,  b,  and  c  is 
found  in  every  one  of  the  expressions,  and  a^,  b^,  and  c  are  the  hi<,diest 
powers  of  a,  b,  and  c  by  which  all  of  the  given  expressions  can  be  divided 
exactly. 

Ex.  3.    Find  the  H.  C.  F.  of  8  a^b^c  and  12  a%^d. 

Observe  that  the  greatest  common  divisor  of  the  numerical  coefficients  8 
and  12  is  4.     The  literal  parts  have  the  highest  common  factor  a%^. 

Accordingly  the  highest  common  factor  sought  must  be  the  product  of 
the  greatest  common  divisor,  4,  and  a%^ ;  that  is,  4  a%^. 

7.  To  find  the  H.  C.  F.  of  two  or  more  monomial  ex- 
pressions : 

Construct  a  term  containing  every  letter  and  'prime  numerical 
factor  which  is  common  to  all  of  the  given  expressions^  taking  each  to 
the  lowest  power  vihich  is  found  in  any  one  of  them, 

8.  It  should  be  observed  that  no  mention  is  made  in  the  defi- 
nition of  the  highest  common  factor  of  that  factor's  numerical  value. 


230  FIRST   COURSE  IN  ALGEBRA 

The  numerical  value  of  the  highest  common  factor  of  two  given  ex- 
pressions is  not  equal  to  the  arithmetic  greatest  common  divisor  of 
the  numerical  values  of  the  given  expressions  when  numerical  values 
are  given  to  the  letters. 

Hence,  numerical  checks  cannot  be  used  for  highest  common 
factors. 

9.  There  is  no  fundamental  connection  between  the  ideas  of 
greatest  common  divisor  in  arithmetic  and  highest  common  factor 
in  algebra.  The  word  "highest,"  used  in  the  definition  of  the  alge- 
braic highest  common  factor,  refers  simply  to  the  degree  of  the 
divisor,  either  with  reference  to  a  particular  letter  or  with  reference 
to  all  of  the  letters  in  a  tenn.     (See  §  5.) 

The  word  "  greatest "  is  used  in  the  definition  of  the  greatest 
common  divisor  of  two  or  more  numbers  in  arithmetic,  because  the 
greatest  common  divisor  is  the  greatest  divisor  by  which  two  or 
more  given  expressions  can  be  divided  without  remainder. 

10.  It  does  not  follow  that  one  expression  represents  a  greater 
number  than  another  because  it  contains  higher  powers  of  one  or 
more  letters. 

E.  g.  The  value  of  a^  is  numerically  less  than  that  of  a  when  a  is  posi- 
tive and  less  than  unity.  The  value  of  a^  is  equal  to  that  of  a  when  a  is 
unity,  and  is  numerically  greater  than  a  for  all  values  of  a  which  are 
neither  unity  nor  numerically  less  than  unity. 

For,  when  «  =  ^.      o^^  =  i ;  that  is,  a-  <  a. 

a  =  1,  a^  =  1 ;  that  is,  a^  =  a. 
a  =  3,  a^  =  9 ;  that  is,  a^  >  a. 
a  =  —  2,  ^2  =:  4 ;  that  is,  a^  >  a. 

Exercise  XIII.     1 

Find  the  highest  common  factor  of  the  expressions  in  each  of  the 
following  groups: 

1.  ax\  af,  az\  6.  ah\  a%d,  aWe. 

2.  a'hc,  ab\  abc\  7.  xhfz\  x'ij'z\  xhfz\ 

3.  x^y\  x^fzw,  y^z'w,  8.  10  a^  and  25  a». 

4.  ^^6%  ^Va,  Mb.  9.  12 ^'c  acd  X^b^. 

5.  ab\m,  a^cmy,  ab\z,  10.  l^abc  and  24  W, 


HIGHEST   COMMON   FACTOR  231 

11.  56«4/;and  Ua'b\  15.  6abd\  18  a^b^e,  '60aW(Pe\ 

12.  42  ^/y  and  12  xy.  16.  Sa^bd,  21  ab^d^e,  da%de\ 

13.  11  a%  and  S'Sab^c^d.  17.  2abx,  10  ahjx,  Sax^z. 
14  100rt'6Vand55a*^>»c.  18.  4a!7/<:^  12a;y;^,  lGxyz\ 

19.  20xyzWj  5xi/z%  li)xi/zw. 

20.  35  a^i^crf,  7  abed,  49  a'^6 W. 

21.  5a*6c»,  10  ab\  35(i^b. 

22.  4a;y,  6a3y;3^  lOccy^,  12  xYz. 

23.  40a»/;c,  24  a^^^  Uabz,  88  a%. 

24.  14  ay,  21  icV>  42  ic/^,  35icy<^2. 

25.  3«a;y^,  12£c/7w,  18'^i/z^  Sxfz. 

26.  SOajy^^w;,  125icy5''M;,  lOOxYz'w,  UxYz. 

27.  18  ar^y^s^w,  27  x}/zw\  81  ar^y^;^;^  45  xyH^w. 

11.  When  two  given  expressions  are  polynomials,  the  degree  of 
the  highest  common  factor  is  usually  reckoned  with  respect  to  some 
particular  letter. 

Highest  Common  Factor  of  Polynomial  Expressions 

12.  If  the  given  expressions  can  be  readily  factored,  we  may 
obtain  the  highest  common  factor  by  inspection  as  follows  : 

Obtain  the  pi' line  factors  of  each  of  the  given  expressions.  Write 
the  product  containing  each  of  the  prime  factors  common  to  all  of  the 
given  expressions,  taking  each  factor  to  the  lowest  power  which  is  found 
in  any  one  of  them. 

Ex.  1.    Find  the  H.  C.  F.  of  «2  +  2 a6  +  h\  a?'  -  b%  and  Sab +  3 h\ 
The  work  may  be  arran^jed  as  follows  : 

a2  +  2  ah  +  h'^={a-\-  b)^ 

The  only  factor  which  is  common  to  all  of  the  expressions  is  the  first 
power  of  a  +  b. 

Hence  the  highest  common  factor  is  a  +  6. 

Ex.  2.   Find  the  H.  C.  F.  of  a:2  +  11  x  +  30,  a;^  +  3a:  -  10,  and  4  a:  +  20. 

We  have  a:^  +  11  a:  +  30  =  (x  +  5)(a;  +  6) 

x^-{.    3a:- 10=(a:  +  5)(a;-2) 
4a: +  20  =  4  (a: +  5) 
Hence  the  H.  C.  F.  is  a:  +  5. 


232  FIRST  COURSE  IN  ALGEBRA 

Ex.  3.    Find  the  H.  C.  F.  of  7  a^b%a  +  h)9(a  -  hy(x  +  y)  and 
6a-^6»(a  +  hy\a  -  h)\x  -  y). 

The  factors  a^,  b^^  (a  +  b)\  and  (a  —  6)'»,  and  no  others,  are  common  to  both 
exprcRsions. 

Hence  the  H.  C.  F.  is  their  product,  a%^{a  +  b)\a  -  bf. 


Exercise  XIII.     2 

Find  the  highest  common  factor  of  the  expressions  in  each  of 
the  following  groups  : 

1.  (a  +  Oy\  (a  +  by.  5.  x"  -  l&,  x"  -  d  x  +  20. 

2.  (x  +  i/f,  x'  -  /.  6.  x^+  5x+  G,x^+e,x-^  9. 

3.  5a(7»- w),  15a\m^  —  ?i^.       7.  «'- 9«,  a^—lla-^  18. 

4.  (or  -  2)^  a^  -  2a.  8.  a'^  +  2ab  +  b\  (a  +  b)\ 
9.  a^  -b^,a-b,a^-2ab  +  b\ 

10.  2 a  —  4y,  a^  —  4^,  ar^  —  4 a;?/  +  4 ?/^. 

11.  3a6  +  36,  2aic+  2x,  Hah  +  Haz. 

12.  a  +  6,  a*  +  2a6  4-  h^\  a*  +  6». 

13.  ar*  +  8a;+  15,  jb^  — a;—  12,  «' +  6a!  +  9. 

14.  xz  +  xw  —  yz  —  yw  and  a:^  —  y"^. 

15.  «*  —  y\  «*  +  /,«  +  3^. 

16.  4f/'  +  20a  4-  25,  2a  +5,  8a*  +  125. 

17.  3a2+ 7a6-206^  a^-  166^  a  +  46. 

18.  2ac-^^ad.  —  2bc-^bdQ,ndi^c'—^oP. 

19.  a^  +  3a  —  10,  a^  +  6a  +  5,  a'  +  2a  —  15. 

20.  2a;*  +  a;-  6,  ar^  +  3  a;  +  2,  a;'  —  4. 

21.  2qi?  —  xy  —  ?/^,  Qi?  —  y^.,  x^  —  2xy  +  y\ 

22.  56a*- 126a,  2a  — 3,  4^2  _  12a  +  9. 

23.  Sla^— 72a6+  Ub\  9a  -  4  6,  81  a^—  16  6^. 

24.  xz  +  xw  +  yz  -\-  yyj,  a^  +  2xy  +  y%  ax  +  bx  +  ay  +  by. 

25.  (a  -  6)*,  a*  -  2  a^b^  +  b\  a»  -  a%  -  ab"-  +  b\ 

13.  The  term  "  greatest  common  divisor  "  is  not  appropriate  when 
applied  to  algebraic  expressions. 

When  numerical  values  are  assigned  to  the  letters  appearing  in 
two  algebraic  expressions  and  also  to  the  letters  appearing  in  their 
highest  common  factor,  it  may  happen  that  the  value  represented 


HIGHEST  COMMON   FACTOR  233 

by  the  highest  common  factor  is  not  the  greatest  common  divisor  of 
the  values  represented  by  the  given  expressions. 

E.  g.  The  expressions  x^  -{-  I  and  x  +  I  have  no  highest  common  factor 
and  are  alj^ehraically  prime.  Yet,  if  3  be  substituted  lor  x,  x'^  -\-  I  becomes 
10,  and  a:  +  1  becomes  4.     The  greatest  common  divisor  of  10  and  4  is  2. 

The  highest  common  factor  of  a:^  _^  7  a;  _|_  j  2  and  x^  -{-  IO2;  +  21  is  a:  +  3. 
If  X  is  given  some  particular  value,  such  as  11,  then  the  expression 
x^  4-  7  a:  +  12  becomes  210  and  the  expression  x^  +  10  a:  +  21  becomes  252, 
while  the  highest  common  factor,  x  +  3,  of  the  algebraic  expressions,  becomes 
14.  The  greatest  common  divisor  of  the  numerical  values  210  and  252, 
however,  is  42,  not  14,  which  was  obtained  by  substituting  11  for  x  in  the 
algebraic  highest  common  factor  x  +  3. 

If,  however,  4  be  substituted  for  x,  the  values  of  these  expressions  and 
their  highest  common  factor  are  56,  77,  and  7,  respectively.  The  value,  7, 
obtained  from  the  highest  common  factor,  is  in  this  case  the  greatest  common 
^divisor  of  the  numerical  values  56  and  77  which  are  represented  by  the 
algebraic  expressions. 

Highest  Common  Factor  of  Polynomials 

14.  When  two  integral  functions  of  some  common  letter,  x,  cannot 
be  readily  factored  by  inspection,  the  process  for  finding  the  highest 
common  factor,  or  of  showing  that  the  functions  are  prime  to  each 
other,  may  be  made  to  depend  upon  the  following  principles  : 

Principle  I.  If  an  integral  expresssion  be  divisible  ivithout 
remainder  by  another  integral  expression  which  is  of  the  same  or  of 
lower  degree  with  reference  to  some  common  letter  of  arrangement^  the 
expression  used'  as  divisor  is  the  highest  common  factor  of  the  two 
expressions.  » 

For,  if  the  divisor  be  contained  without  remainder  in  the  dividend,  it  is  a 
factor  of  the  dividend,  and  hence,  by  definition,  must  be  the  highest  common 
factor. 

Principle  II.  The  highest  common  factor  {if  there  be  one)  of  two 
integral  polynomial  expressions  is  also  the  highsst  common  factor  of 
the  divisor  and  the  integral  remainder  obtained  by  dividing  one 
expression  by  the  other. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time. ) 

Let  D  and  d  represent  any  two  integral  expressions  arranged  according 
to  descending  powers  of  some  common  letter,  x,  the  degree  of  the  divisor 


234  FIRST  COURSE   IN  ALGEBRA 

d  being  not  higher  than  that  of  the  dividend  D  with  reference  to  the  letter 
of  arrangement. 

Let  both  the  quotient  Q  and  the  remainder  12,  obtained  by  dividing  the 
dividend  D  by  the  divisor  d,,  be  integral. 

(1)  1/  D  and  d  have  a  highest  common  factor ^  denoted  by  h,  then  d  and  B 
will  luive  a  highest  common  f odor  which  is  the  same  expression,  h. 

If  all  of  the  terms  of  either  of  the  given  expressions  D  or  d  have  common 
numerical  or  monomial  factors,  these  should  first  be  removed  by  division. 
If  the  common  factors  thus  removed  have  a  highest  common  factor  this 
should  he  set  aside  as  a  factor  of  the  required  highest  common  factor  of  the 
two  given  expressions  D  and  d. 

Accortlingly  we  shall  assume  in  the  following  proof  that  the  given  ex- 
pressions D  and  d  have  neither  numerical  nor  monomial  factors  common  to 
all  of  their  terms. 

Representing  by  h  the  highest  conmion  factor  of  the  dividend  D,  and  the 
divisor  d,  we  may  write 


d  =  nh.) 


0) 


The  factors  m  and  n  of  the  right  members  of  identities  (1)  must  be  prime 
to  each  other,  otherwise  h  would  not  be  the  highest  common  factor  of  D 
and  d. 

Since  the  dividend  D  is  equal  to  the  divisor  d  multiplied  by  the  quotient 
Q,  plus  the  remainder  R,  we  may  write 

I)  =  dQ  +  R.  (2) 

Substituting  in  (2)  the  values  for  D  and  d,  from  (I),  we  obtain 
mh  ='/nhQ  +  R. 
Hence  mh  —  nhQ  =  R. 

Or,  h(m-.nQ)=R.  (3) 

Since  both  members  of  the  last  identity  are  integral  expressions,  and  h  is 
a  factor  of  the  first  member,  it  must  be  a  factor  of  the  second  member  also. 
That  is,  h  must  be  a  factor  of  R. 

Since  h  is  the  highest  common  factor  of  D  and  d,  it  follows  that  h  must 
be  a  factor  of  d. 

We  have  shown  by  the  reasoning  above  that  ^  is  a  factor  of  the  remain- 
der R  also. 

Hence  d  and  R  must  Jiave  a  com,mon  factor  which  is  at  least  h.    ^ 

We  will  show  that  d  and  R  have  no  factor  in  common  other  than  h. 


HIGHEST   COMMON   FACTOR  235 

An  expression  as  a  whole  is  exactly  divisible  by  another  expression  if  its 
terms  are  separately  divisible  by  the  other  expression. 

The  highest  common  factor  of  d  and  B  must  be  a  divisor  of  the  expres- 
sion dQ  +  R,  for  it  is  contained  in  each  of  the  terms  dQ  and  B. 

It  follows  from  the  identity  D  =  dQ  +  R{2)  that  every  factor  of  the 
second  member  dQ  -\-  B  must  also  be  a  factor  of  the  first  member  D.  Such 
a  factor  which  is  common  to  both  members  of  the  identity  D  =  dQ  -}■  B 
must  be  a  factor  of  both  d  and  D. 

Hence  the  highest  common  factor  of  d  and  B  must  be  a  factor  also  of  D, 
that  is,  t)ie  highest  common  factor  of  d  and  B  cannot  exceed  h  which  is  the 
highest  common  factor  of  d  and  D. 

It  follows  from  the  reasoning  above  that  tlie  highest  common  factor  of  two 
given  expressions  is  preserved  in  tJie  integral  remainder  {if  there  be  one)  after 
division. 

If  this  remainder  be  used  as  a  new  divisor  and  the  divisor  first  used  be 
taken  for  a  new  dividend,  the  principle  will  hold  as  before,  and  the  highest 
common  factor  will  be  carried  over  again,  and  will  be  found  in  the  second 
remainder  (if  there  be  one)  after  division, 

(2)  //,  however,  D  and  d  he  prim^  to  each  other,  tlien  d  and  B  will  also  he 
prime  to  each  other. 

If  the  dividend  D  and  the  divisor  d  have  no  factors  in  common,  that  is, 
if  they  are  prime  to  each  other,  it  follows  that  d  and  B  can  have  no  factors 
in  common. 

This  is  because  all  of  the  factors  which  are  common  to  d  and  B  are  con- 
tained in  each  of  them,  and  from  the  identity  jD  =  dQ  +  B  (2),  it  follows 
that  such  ftictors  must  be  contained  also  in  D. 

Hence  it  appears  that  if  d  and  B  have  any  common  factor  it  will  con- 
tradict our  assumption  that  D  and  d  are  prime  to  each  other. 

From  the  reasoning  above  it  appears  that  if  D  and  d  have  no  highest 
common  factor  the  divisor  d  and  the  remainder  B  will  have  no  highest  common 
factor. 

15.  Development  of  the  Process.  From  the  principle  above 
it  appears  that,  instead  of  examining  two  given  expressions  to  de- 
termine whether  or  not  they  have  a  highest  common  factor,  we 
may  divide  one  expression  by  the  other  and  examine  for  a  highest 
common  factor  the  divisor  and  the  remainder  resulting  from  the 
division. 

16.  The  "  division  "  may  be  carried  out  according  to  descending 
powers  of  some  letter  of  arrangement  until  a  remainder  is  obtained 
which  is  either  a  constant  or  an  expression  of  lower  degree  than  the 


236  FIRST  COURSE   IN   ALGEBRA 

divisor  with  reference  to  the  common  letter  of  arrangement  accord- 
ing to  which  the  division  is  being  performed. 

If  the  remainder  is  an  expression  containing  the  letter  according 
to  which  the  division  is  being  performed,  it  may  be  taken  as  a  new 
divisor,  and  the  divisor  previously  used  may  be  taken  as  a  new 
dividend. 

It  follows  that  by  repeating  this  process  we  must  sooner  or  later 
reach  a  stage  at  which  : 

Either  the  remainder-divisor  is  contained  exactly  in  the  corre- 
sponding remainder-dividend, 

Or,  the  last  remainder  is  a  constant  free  from  the  letter  of  arrange- 
ment. In  this  last  case  the  process  of  "  division  "  must  stop,  that 
is,  become  "  inexact "  at  this  point. 

17.  In  case  the  last  remainder-divisor  is  contained  in  the  cor- 
responding remainder-dividend,  it  must,  by  Principle  I,  §  14,  be  the 
highest  common  factor  of  "  itself  "  and  the  corresponding  dividend. 
Hence,  it  must  be  the  H.  C.  F.  of  all  previous  pairs  of  corresponding 
remainder-divisors  and  remainder-dividends,  and  conseciuently  must 
be  the  H.  C.  F.  of  the  original  divisor  and  the  original  dividend. 

18.  If,  however,  the  last  remainder  is  different  from  zero,  and  is 
a  constant  free  from  the  letter  of  arrangement  according  to  which 
the  "division"  is  being  performed,  then  the  last  remainder-divisor 
and  remainder-dividend  have  no  highest  common  factor,  and  accord- 
ingly the  preceding  pairs  of  remainder-divisors  and  remainder-divi- 
dends have  "none.  Therefore  the  original  functions  have  no  highest 
common  factor  ;  that  is,  they  must  be  prime  to  each  other. 

19.*   Denoting  the  original  expressions  by  D  and  d,  and  the  successive 

intej^ral  quotients  by  Q^,  Q2,  Qg, ,  and  the  successive  remainders  by 

R^,  -K2,  i?8, ,  we  may  indicate  the  process  as  follows : 


Indicated  Process 


Pairs  of  expressions  which  all 
have  the  same  H.  C.  F. 

d)  D        o d,    B 

R^  \£_ Rx,d. 

R2IR1 ^2,^1- 

-^3  )  -^2 -^8'  -^2* 

etc.  etc. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


HIGHEST  COMMON   FACTOR  237 

In  this  process,  I)  =  d  (J^  +  i?i 

d  =  i?i(?2  +  ^2 
i?2  =  ^^8^4  +  ^4 


20.  The  process  above,  for  finding  the  highest  common  factor  of 
two  integral  functions,  consists  in  substituting  for  the  original  pair 
of  functions  a  second  pair,  for  these  a  third  pair,  and  so  on,  all  pairs 
of  functions  having  the  same  highest  common  factor. 

Ex.  1.    Find  the  H.  C.  F.  of  a:«  -  x^  +  2  and  .t«  -  2  a-2  +  3. 

In  order  to  make  the  arrangement  of  the  process  correspond  closely  with 
that  shown  in  the  next  section,  §  21,  we  shall,  throughout  the  work,  write 
the  divisor  at  the  left  and  the  "  quotient "  at  the  right  of  the  dividend. 

The  first  stage  of  the  work  is  carried  out  below  : 

DiviBor  Dividend        Quotient 

x8  -    g2  +  2  First  Stage 

First  Remainder    ...       —      ic'-^  +  I 

The  remainder  from  the  division,  —  x^  +  l,  is  of  lower  degree  with  refer- 
ence to  X  than  the  divisor  x^  —  x^  -\-  2. 

By  Principle  II.  §  14,  we  know  that  the  H.  C.  F.,  if  there  be  one,  of  the 
divisor  and  dividend  must  be  contained  as  a  factor  of  the  remainder,  —  .r^  +  1, 
and  must  be  the  H.  C.  F.  of  this  remainder  —  x^  +  I  and  the  divisor 
aH»  -  a:2  +  2. 

For  the  second  stage  of  the  work  we  will  use  this  first  remainder  as  a 
new  divisor  and  the  first  divisor,  x^  —  x'^  +  2,  as  a  new  or  second  dividend, 
as  follows : 

Original  Divisor    Quotient 
First  Remainder-Divisor    —  X^  +  I)  X^  —  X^  -\-  2  )  —  X  -\-  I 

X^  —  X 

—  x'^  +  x  4-  2  Second  Stage 

-x^  +1 

Second  Remainder +  a;  +  1 

This  second  remainder,  x  +  1,  must  contain  the  H.  C.  F.  of  the  original 
expressions.  The  work  of  the  third  stage  of  the  process  may  he  carried  out 
hy  taking  this  second  remainder,  a:4-  1,  as  a  new  divisor  and  the  corre- 
sponding divisor,  —  x'^  -\-  1,  as  a  new  or  third  dividend,  as  follows : 


238  FIRST  COURSE  IN  ALGEBRA 

Second  Remainder-         First  Remainder-  Quotient 

Divisor  Dividend 

H.  C.  F.  sought      ....     a:  +  1  )  _  ar'i  +  1  )  -  a:  +  1 

-  .r^  -  X 

+  a;+  I 

Third  Stage  +3^+1 


At  this  stage  of  the  work,  the  divisor  a:  +  1  is  contained  without  re- 
mainder in  the  corresponding'  dividend  —  a:^  +  1. 

Hence,  by  Principle  I,  §  14,  it  must  be  the  H.  C.  F.  of ''itself "  and  the 
corresponding  dividend  —  a^*  -f  1,  and  by  Principle  II,  §  14,  il  must  be  the 
H.  C.  F.  of  all  previous  pairs  of  corresponding  divisors  and  dividends,  and 
hence,  finally,  of  the  original  expressions. 

21.  The  different  steps  of  the  process  may  be  arranged  in  the 
following  compact  oblique  form: 


i                                   a:»-     a:2  +  2 

First  Stage 

:  Urst  Remainder- 
:            divisor 

-     a:«-fl)aH»- 

-a:«-f  2)-a: 

-X 

-  a:2  +  X  -f  2 
-ar2          -f  1 
.  .  .    a:+l) 

+  1 
Second  Stage 

:  Second  Remainder- 
divisor 

H.C.F.  sought . 
Third  Stage 

-  a:2  +  1  )  _  a;  +  1      \ 
x^-x                        \ 

0             i 

22.  Whenever  during  the  process  of  "division"  a  "quotient" 
is  obtained  which  is  not  integral,  we  may  apply  the  following 
Principle:  At  any  stage  of  the  iwocess  of  finding  the  highest  com- 
mon factor,  any  remainder-dividend  or  corresponding  remainder- 
divisor  may  he  7nultiplied  by  or  divided  by  any  number  or  expression 
which  is  not  already  a  factor  of  the  other. 

23.  Caution.  If  at  any  stage  of  the  process  any  common  factor 
is  removed  by  division  from  both  the  dividend  and  the  corresponding 
divisor,  this  common  factor  must  be  set  aside  to  be  used  as  a 
multiplier  of  the  polynomial  highest  common  factor  resulting  from 
the  "division"  process. 


HIGHEST  COMMON  FACTOR  239 

24.  In  practice  it  is  often  convenient  to  remove  numerical  fac- 
tors from  divisors  by  division  and  to  introduce  numerical  factors 
into  dividends  by  multiplication. 

By  thus  transforming  the  terms  of  the  divisors  and  dividends 
it  is  possible  to  avoid  fractional  *'  quotients  "  at  any  stage  of  the 
process. 

Ex.  2.  Find  the  H.  C.  F.  of  3x4  +  2 x^  +  4 x^  +  a:  +  2  and  2x*y-\-5  x^y 
+  5  xhj  +  3  xy. 

We  will  first  remove  the  common  monomial  ftictor  xy  from  the  terms  of 
the  second  expression  by  division,  a.s  follows: 

2  a:*j/  +  5  x*?/  +  5  x^y  +  3  xi/  =  xj/  (2  x8  +  5  x2  -i-  5  X  +  3). 

Neither  of  the  factors  removed  (x  nor  y)  is  a  common  factor  of  all  of  the 
terms  of  the  first  expression  3x4+  2x*  +  4x*  +  x-f-2.  Hence,  neither  x 
nor  y  can  be  contained  as  a  factor  of  the  highest  common  factor  of  the 
given  expressions. 

Accordingly  the  factors  x  and  y  which  were  removed  by  division  from 
the  second  expression  may  be  neglected. 

The  expressions  2  x^  +  5  x'^  +  5  x  +  3  and  3  x^  +  2  x^  +  4  x^  +  .r  +  2  may 
now  be  used  as  divisor  and  dividend  respectively  to  find  the  desired  highest 
common  factor. 

The  fractional  "quotient,"  |x,  which  would  be  obtained  by  dividing  the 
first  term  3x4  ^f  ^1,^  dividend  by  the  first  term  2x*  of  the  divisor,  may  be 
avoided  by  applying  the  princii)le  of  §  22,  that  is,  by  multiplying  all  of  the 
terms  of  the  dividend  3  x4  +  2  x*  +  4  x^  +  x  +  2  by  2. 

Since  the  factor,  2,  thus  introduced  into  the  dividend  by  multiplication, 
is  not  also  a  factor  of  all  of  the  terms  of  the  divisor,  the  value  of  the  highest 
common  factor  sought  will  not  be  aftected. 

The  steps  of  the  process  are  shown  below  : 


Modified  Divisor. 

Original  Dividend. 

Place  for  Quotients. 

2x8  +  5x2 

+  5x  H 

1-3)3x4 

2(T 

+    2.A»+    4x2+       a;  + 

o  avoid  fractional  coefficients). 

2 

6x4 

+    4x3+    8a;2_^    2x  + 

0 

3x 

6x4 

+  15x8+  i5a;2_{_    9  a; 

-Ux^-    7x2-    7a._^    4 
2  (To  avoid  fractional  coef.). 

First  Stage. 

-22x8-  14x2-  14a;  4. 

8) 

-11 

-  22  x8  -  55  x2  -  55  X  -  ; 

33 

Numerical  factor  41  removed, 
aince    it    cannot    belong    to    the 
H.  C.  F.  sought. 

41  )  +41x2 +  41  x  + 41 

X2+         X+      1 

240  FIRST  COURSE  IN  ALGEBRA 

The  remainder  a:*  +  x  +  1  is  of  lower  degree  with  reference  to  x  than  the 
corresponding  modified  divisor  2  a;*  +  5  a;''  +  5  x  +  3. 

The  process  may  now  be  continued  by  using  this  remainder  x^  +  ;r  +  1 
as  a  new  divisor,  and  the  first  divisor,  2x*  +  5ar2  +  5x  +  3,  asa  new  divi- 
dend, as  follows  : 

RemaiBder-  Modified  divisor  Place  for 

divisor  used  as  divideud  quotients 

H.  C.  F.  sought    a:2  +  a:+l)2x3  4-5x2  +  5a:  +  3  )  2a; 

2  a:«  +  2  a:'^  +  2  a: 

3  )  3  x''  +  3"a:  +  3    Secoud  Stage 

...         x-2  +     x  +  1  )  I 
x2  4-     X  4-  1 


Numerical  factor  removed 


The  numbers  written  in  the  place  for  quotients  cannot  be  considered 
quotients  in  the  ordinary  sense,  owing  to  the  introduction  and  rejection  of 
factors  during  the  work. 

The  division  becomes  exact  when  x*  +  x  +  1  is  used  as  a  divisor. 
Hence,  by  Principles  I  and  II,  §  14,  it  must  be  the  H.  C.  F.  of  the  given 
expressions. 

25.  It  will  be  seen  in  the  work  above  that,  each  time  a  remainder 
is  taken  as  a  new  divisor,  the  first  term  is  contained  an  integral 
number  of  times  in  the  first  term  of  the  corresponding  dividend. 
It  will  be  found  that  this  very  seldom  happens  in  practice. 

26.  It  should  be  observed  that  the  successive  "divisions"  per- 
formed when  carrying  out  the  process  for  finding  the  highest  com- 
mon factor  are  not  commonly  "  divisions  "  in  the  ordinary  sense, 
since  at  different  stages  of  the  work  we  may  introduce  or  remove 
factors  from  either  the  dividend  or  the  divisor. 

27.  For  this  reason  the  expressions  written  in  the  places  of  quo- 
tients are  not  "quotients"* in  the  ordinary  sense,  and  since  in  the 
result  we  are  not  at  all  concerned  with  the  "  quotients,"  we  may 
neglect  writing  them  altogether. 

28.  It  will  be  seen  that  in  the  "  oblique "  arrangement  of  the 
work  used  in  the  examples  of  §§  21,  24,  it  is  necessary  to  copy 
again  each  divisor  when  it  is  used  as  a  new  dividend.  Further- 
more, when  arranged  in  compact  form,  the  work  tends  to  extend 
downward  in  an  oblique  direction,  toward  the  right. 


HIGHEST  COMMON   FACTOR 

These  objections  may  be  overcome  by  adopting  a  vertical  ar- 
rangement for  the  work. 

The  contrast  between  the  oblique  and  the  vertical  arrangements 
may  be  shown  by  carrying  out  the  work  of  the  example  of  §  24  in 
vertical  arrangement. 

The  given  expressions  are  separated  by  vertical  lines,  as  shown 
below,  and  the  "  quotients  "  are  placed  in  the  side  columns  nearest 
their  dividends. 

The  divisors  will  be  found  sometimes  on  the  left  and  sometimes 
on  the  right  of  the  corresponding  dividends. 


241 
ar-1M 

ents      " 


:      Place                                                  Dividend                                                                      Place      : 

for                                                        and                                                                             for 
;  Quotients                                              Divisor                                                                    Quotients  [ 

2x8 +  5x2  + 5a: +  3 

3a;*  +    2a:8+    4x^-\-      a:  +    2 
2     (To  avoid  frac.  coef.) 

3a:          1 

First    i 
Stage   i 

-11         i 

2x 

1 

2aH»  +  2a:2  +  2a: 

3)3a:2  +  3a:  +  3  \ 
a:2+    a:+l   i 
a:2+    a:+l  ; 

0  ; 

Oa:*+    4a:8+    8a,-2  +    2a:  +    4 
6a:*+15x«+15a:2+    9a: 

-lla:8-    'jx2_    7a:+    4 
2      (To  avoid  frac.  coef.) 

-  22a:8-14a:2-14a:+    8 
-22a:8-55a:2-55a:-33 

41)41a:2  +  41a:  +  41 

a:2+      a:+    1 
H.  C.  F.  sought. 

Second ; 
Stage   : 

29.  In  finding  the  H.  C  F.  by  the  long  "  division  "  process,  the 
simple  numerical  or  monomial  factors  must  first  be  removed  by 
division  from  the  terms  of  the  expressions. 

The  H.  C.  F.  (if  there  be  any)  of  these  factors  thus  removed 
must  be  set  aside  to  be  used  as  a  multiplier  of  the  polynomial  high- 
est common  factor  resulting  from  the  "  division  "  process. 


Ex.  3.    Find  the  H.  C.  F.  of  2a%^  -  4a*b*  +  4a%^ 
2a*62  +  a868. 

16 


2  a^b^  and  a^b  — 


242 


FIEST  COURSE  IN  ALGEBRA 


a%)a^b  +  2a*b^  +  a%» 
a2  -2ab    +  b'^ 

2 a^fts  )  2 a^b»  -  4 a*b*  +  4a%^  - 
o8     -2a^b  +2a62  - 
flS     _2a26  +    a62 

2a266 
6» 

a 

a 

a^  —    ab 

:  -b 

-  ab    +  b'^  ; 

-  ab    +  b-^  "" 

^»2  )  ab^  - 

68 

H.C.F.  of  modi-        a    - 
fied  expressions. 

6 

0 

The  H.  C.  F.  of  a%  and  2  a'^ft^  which  were  removed  by  division  at  the 
beginning  of  the  work,  is  a%.  Hence  the  H.  C.  F.  sought  is  the  product 
of  a%  and  the  polynomial  highest  common  factor  (a  —  b).  That  is,  the 
highest  common  factor  is  a%{a  —  6). 

30.  The  process  for  finding  the  highest  common  factor  has  the 
peculiarity  of  not  only  furnishing  the  highest  common  factor  (if  it 
exists),  but  also  of  indicating  when  there  is  none. 

31.  To  find  the  H.C.F.  of  two  integrral  functions  (if  it 
exists)  we  may  proceed  as  follows  : 

AJier  hainng  first  removed  all  common  monomial  factors,  treat  the 
modified  expressions  as  dividend  and  divisor. 

If  the  degrees  of  the  expressions  are  the  same  with  reference  to  the 
common  letter  of  arrangement  either  expression  may  he  used  as 
divisory  hut  if  the  degrees  are  not  the  same  then  the  expression  oj 
lower  degree  must  he  used  as  divisor. 

Continue  the  process  of  "  division "  until  the  degree  of  the  re- 
mainder tvith  reference  to  the  letter  of  arrangement  is  at  least  as 
low  as  the  degree  of  the  divisor. 

Using  this  remainder  as  a  new  divisor,  and  the  first  divisor  as  a 
new  dividend,  the  process  may  he  repeated  until  finally  either  the 
"  division "  becomes  exact,  —  in  which  case  the  last  divisor  used, 
multiplied  hy  such  common  factoi'S  as  may  have  been  removed  at  the 
heginning  of  the  work,  is  the  H.C.F.  sought,  —  or  until  a  remainder 
is  obtained  which  is  free  from  the  letter  of  arrangement,  in  which 
case  no  H.C.F.  exists. 

During  the  process  the  separate  dividends  and  divisors  may  be 
divided  hy  m-  multiplied  by  stich  numerical  or  literal  factors  as  are 
necessary  to  avoid  fractional  coefficients  in  the  "  quotients "  in  the 
course  of  the  "  division." 


HIGHEST   COMMON  FACTOR  243 

Exercise  XIIL   3 
Find  the  H.  C  F.  of  each  of  the  following  groups  of  expressions  : 

1.  a?  +  2i^-8x—U,x^  +  Sx'-8x-24:. 

2.  z^-5z  +  4,z^-5z'  +  4:. 

3.  «*~a2-5a  — 3,  a»-4«'- Ua  — 6. 

4.  ^»  +  2b^-rdb+  10,  b^  +  b^  -  106  +  8. 

5.  4c*- 3c2-  24c- 9,  8c«  — 2c2  — 53c  — 39. 

6.  2d'  —  5d+2,V2d^  —  S(P  —  Sd+2. 

7.  /+ 2^2+ 2^+  1, /- 2^— 1. 

8.  h*  -  2  A^'  +  1,  A*  -  4^«  +  G/^•■'  -  4/i  +  1. 

9.  2s'' +  '6sH-^,is^  +  st'-t\ 

10.  w*  —  2w^  +  w,2w*  —  2w*  —  2w-'2. 

11.  2ic'-5iB  +  2and2a;»-3aj2-8a;+ 12. 

12.  4a»  -  a'^^  -  ab'' -  b b^  and  7  a»  +  4^26  +  io^)'*  -  3  ^>». 

13.  3y  -  4/  +  2«/^  +  ?/  —  2  and  3^*  +  8/  -  b y^  —  Gij. 

14.  6«  +  3  /;'  4-  4  6  +  2  and  2  6»  +  /^'-^  +  1. 

15.  2c*+ 9c«+  14c4- 3and  3c'+  15c«  + 5c2+  10c  + 2. 

16.  ^d^—'6d''  —  SSd+  9  and  6^*+  (^cP  -42d'—  ISd. 

17.  3A«- 9/^=*+ 2U- 63and2^*+  19^'+ 35. 

18.  Qa^  -ea^b  +  2ab^  -  2b^  and  ^(i"  -bab-{-  b\ 

19.  m^  +lm^  +lm}—\b  m  and  2  m*  —  4  m^  —  26  ?w  +  220. 

20.  4wHl4w*  +  20?j*+70w2and8;i'4-28w'  — 8w^  — 12w'+56w'. 

21.  2  w;*  —  2  i^;*  +  4  w?^  +  2  w;  +  6  and  3  i^?*  +  6  m;*  —  3  z^  —  6. 

22.  ic*  +  ic»  —  9  ar^  —  3  a;  +  18  and  aj^  +  6  a;2  -  49  a;  +  42.  ■ 

23.  8s*—  6;s«+  3s^  — 3s+  1  and  18s«-  3s*-  15s  +  6. 

24.  64/  —  3/A*  +  5^A*  and  20/  —  3/y%  +  ^*. 

25.  2s*-  7s*  +  9s*  — 85-  5  and  6s*—  11  s"-  16 s"^+  155. 
The  H.  C.  F.  of  three  or  more  expressions  may  be  obtained  by  first 

finding  the  H.  G.  F.  of  any  two  of  them^  then  the  H.  C,  F.  of  this 
result  and  a  third  expression^  and  so  on. 

This  is  because  any  factor  which  is  common  to  three  or  more  expressions 
must  be  a  factor  of  the  H.  C.  F.  of  any  two  of  them. 

26.  ?w*  +  m  —  6,  ?w*  —  2  7W*  -  w  +  2  and  w'  +  3  w*  —  6  w  —  8. 

27.  6«  4- 7  6*+ 56  -  1,  36*+ 56*  + 6- land  3  6«- 6*- 36+1. 

28.  a*  -  6a*  +  11  a  —  6,  «*  -  9a*  +  26a  -  24  and 

a* -8  a*  +  19  a-  12. 


244  FIRST  COURSE   IN  ALGEBRA 

29.  c»  -  9c2  +  26c  -  24,  &  -  lOc^  +  31  c  -  30  and 

c«-  llc2+  38  c -40. 

30.  d^  -  1,  d''  -  d^  -  d  -  2  &nd  d^  -  2d^  -  2d  -  3. 

31.  c*  -  c«  —  c^'  —  2  c  —  1,  c*  +  c»  +  c^  +  1  and 

c^  +  2c*  +  c8  +  c2- 1. 

32.  k'  -  1,  /i'  -h'-  h^-h  -  1  and  A«  -  h' +  h' -  h' +  h^-h. 

33.  5a:*+ 7.c*-7a;'-6aj  + 4,  5 a;* +12a^- 15 ar^- 14a; +12and 

5a;*-3a;»+  4ar'+  8a;-8. 

34.  46-'  +  46«-  13^?=*-  15s,  4s»-86'^-75+  15  and 

4s*-26'»-4*''-3s-  15. 


LOWEST   COMMON   MULTIPLE  245 


CHAPTER  XIV 

LOWEST  COMMON  MULTIPLE 

1.  A  coninion  multiple  'of  two  or  more  integral  algebraic  ex- 
pressions is  an  integral  expression  which  may  be  divided  by  each  of 
them  without  remainder. 

2.  The  lowest  coinniou  multiple  or  L.  C  M.  of  two  or  more 
integral  algebraic  expressions  is  the  integral  expression  of  lowest 
degree  which  may  be  divided  by  each  of  them  without  remainder. 

The  lowest  common  multiple  of  two  numbers  or  expressions  which 
are  prime  to  each  other  must  accordingly  be  their  product. 

When  dealing  with  monomials,  the  degree  of  the  lowest  common 
multiple  may  be  reckoned  in  terms  of  several  letters,  while  if  we  are 
considering  integral  polynomial  functions  of  some  single  letter,  say 
X,  the  degree  of  the  lowest  common  multiple  is  determined  by  the 
powers  of  the  common  letter  of  arrangement,  x. 

Lowest  Common  Multiple  of  Monomials  and  Polynomials 
which  can  be  readily  factored 

3.  In  order  to  be  exactly  divisible  by  each  of  the  given  expres- 
sions, the  lowest  common  multiple  of  two  or  more  given  expressions 
must  contain  every  prime  factor  of  each  of  them.  Each  prime  factor 
appearing  in  it  must  be  raised  to  a  power  equal  to  the  highest  power 
of  this  factor  which  is  found  in  any  one  of  the  given  expressions. 

Hence,  to  find  the  lowest  common  multiple  of  ttvo  or  more 
expressions,  construct  an  expression  consisting  of  the  product  of  all 
of  the  different  prime  factors  (numerical  and  literal)  found  in  the 
giveti  expressions,  each  prime  factor  being  raised  to  the  highest  power 
which  is  found  in  any  one  of  them. 

Ex.  1.    Find  the  L.  C.  M.  oi^2a%,  Qa%\  and  l^ahhH. 
We  may  exhibit  the  prime  factors  of  the  numerical  coefficients,  together 
with  the  literal  factors,  by  writing  the  expressions  as  follows: 


246  FIRST   COURSE  IN  ALGEBRA 

32  a%  =  26  •  a^b 
6a»62c  =  2  •3'a%^c 
12  a68c2rf  =  22  •  3  •  aft^c^fi 

The  different  prime  factors  are  found  to  be  2,  3,  a,  b,  c,  and  d,  and  the 
L.  C.  M.  must  be  constructed  by  writing  the  product  of  these  prime  factors, 
each  factor  being  raised  to  the  highest  power  which  is  found  in  any  one  of 
the  given  expressions. 

Hence  the  L.  C.  M.  is  2^  •  3  •  a^b»cH  =  96  a^bhH. 

Ex.  2.    Find  the  L.  C.  M.  of  eab^(a  +  b)^  and  4  rt^^Ca^  -  b^). 
6  ab^(a  +  &)2  =  2  •  3  ab'\a  +  b)^ 
4  a^ia^  -  b-^)  =  2^  •  a%{a  +  b)(a  -  6) 

Hence  the  L.  C.  M.  is 

2^'3'a%\a  +  b)%a-b)=Ua%\a  +  h)^a-h) 

=  1 2  a^b^  + 12  a*6»  -  1 2  a%*  -  1 2  tt^fcs^ 

Ex.  3.    Find  the  L.  C.  M.  of  x^  +  7  a;  +  12,  a;^  -  16,  and  a:*  +  6  x  +  9. 

a;2  +  7a;+  12=  (a:  +  3)(x  +  4) 
a:2  -  16  =  (a;  +  4)(a;  -  4) 

a:2  +  6a;+  9=(z  +  3)(x  +  3) 
Hence  the  L.  C.  M.  is  (x  +  3)2(x  +  4)(a;  -  4)  =  x'  +  6  x^  -  7  a;^  -  96  x  - 144. 

Exercise  XIV.     1 

Find  the  lowest  common  multiple  of  the  expressions  in  each  of  the 
following  groups : 

1.  a,  6,  c.  6.  4  r,  6 .9,  9 1. 

2.  xi/y  yz,  zw.  7.  8  7WW,  lOwzX  \^rnn^. 

3.  rt^6,  6*c,  6-^a.  8.  5F^//,  ^m^xy^  10  n^wy. 

4.  6W,  ^c^  ^c(^.  9.  tf^^  ahc\  abccP. 

5.  ^''^j  ^^ic,  ^w.  10.  14.9£c*«^,  2Sty^Wy  2x^y\ 

11.  6^2^=^,  16 /F,  9^V,  18/^2. 

12.  3  c'cPe',  4  ^£^y^,  5  a^j/V,  6  /;2W. 

13.  ar*  +  5a;  +  6,  ar*  +  6a;  +  8. 

14.  y''-\-2y-  15,  /  —  4^/  +  3. 

15.  ;2'-  15;^+  54,  ;^2-  18  ;^  +  81. 

16.  ab-5b,a^  —  25,  a'  —  10a  +  25. 

17.  w^  —  7i^  m^  —  w^  w'^  +  2  WW  +  w^. 

18.  r  +  s,r^  +  s%T^  +  ^. 


LOWEST  COMMON  MULTIPLE  247 

19.  w"^  —l,w^—  1,  IV— 1. 

20.  h+  l,h  +  2,h^  +  5k  +  Q, 

21.  ^2  -  1,  /  -  1,  ^'  -  f. 

22.  2a2  +  «  —  3,  Sa^  +  a  -  4,  4^2  +  «  —  5. 

23.  5c2+  26c  +  5,  5c2+  31c+  6,  5c2+  36c  +  7. 

24.  cx-{-  dx  +  c?/  +  dy,  x(x  +  2  ?/)  +  ^/^  ^^  —  d^- 

liOwest  Common  Multiple  by  means  of  Highest 
Common  Factor 

4.  If  two  integral  polynomial  functions  of  a  single  letter  cannot 
be  readily  factored  by  inspection,  we  may  find  their  lowest  common 
multiple  by  making  use  of  their  highest  common  factor. 

For,  representing  any  two  integral  expressions,  which  are  not 
prime  to  each  other,  by  A  and  B^  and  denoting  their  highest 
common  factor  by  hy  we  may  write 


A  =  ak,  ) 
B=bh.] 


(1) 


Since  h  represents  the  highest  common  factor  of  A  and  B,  it 
contains  every  factor  which  is  common  to  A  and  B^  and  hence 
a  and  b  must  be  prime  to  each  other. 

The  lowest  common  multiple  of  A  and  B^  represented  by  X, 
must  be  the  lowest  common  multiple  of  the  right  members  ak  and 
bh  of  the  identities  (1). 

Hence  we  have  L  =.  abh. 

The  value  of  the  right  member  abh  remains  unaltered  if  it  be 
successively  multiplied  by  and  divided  by  h. 

Hence  L  =  abh  x  h-r-  h 

_  (ak)(bh) 
-        h 

That  is,  ^-^'  ^^) 

Or^  the  lowest  common  multiple  of  two  integral  expressions  may  be 
found  by  dividing  their  product  by  their  highest  common  factor, 

5.  The  lowest  commcm  multiple  of  two  expressions  may  be  found 
by  dividing  either  one  of  them  by  their  highest  common  factor^  and 
multiplying  the  quotient  by  the  other  expression. 


248  FIRST  COURSE  IN   ALGEBRA 


For,  from  Z  =  ^,  (See  §  4) 

,  .  _  {ah)B 

we  have    L  =  ^~ — » 
n 

or       L  =  (iB. 


Also  from  L  =  -^j 
n 

we  have     L  =  — r-^> 
h 

or       L  =  Ah. 


6.  The  product  obtained  by  multiplying  the  lawest  common  multiple 
of  two  expressions  by  their  highest  common  factor  is  equal  to  the  product 
of  the  two  given  expressions. 

For,  from  the  identity  L  =  —z—* 

h 

we  obtain,  multiplying  both  numbers  by  ^, 

Lh^AB. 

Ex.  1.   Find  the  L.  C.  M.  of  x«  +  5  x^*  +  a:  -  10  and  a;*  +  a:^  _  a;  +  2. 

The  H.  C.  F.  is  fonml  to  Ije  a:  +  2.  The  quotient  obtained  by  dividing 
the  first  expression  a:*  +  5  a;^  +  a;  —  10  by  a;  +  2  is  a:^  +  3  a;  —  5.  Hence, 
the  L.  C.  M.  must  be  the  product  obtained  by  multiplying  this  quotient  by 
the  other  expression,  that  is, 

(aH»  +  a:2  -  X  +  2)(x2  +  3  x  -  n)  =  ar^  ^  4  a.4  _  3  a:8  _  q  a:2  +  H  x  -  10. 

Exercise  XIV.    2 

Find  the  lowest  common  multiple  of  the  expressions  in  each  of  the 
following  groups  : 

1.  «*  +  5a^  +  ha^—ha  —  6  and  ^»  +  6^^+  11  «  +  6. 

2.  6«  +  3  6=^  -  ^  -  3  and  6»  +  4  if;2  +  />  -  0. 

3.  c«  +  c^  -  8  c  —  6  and  2  c*  —  5  c^  -  2  c  +  2. 

4.  ^x^ij—1  aoi^y  —  20 a^xy  and  6 a;^  +  2 ax  —  8 a\ 

5.  2d^-2d^  —  2d—2a,nAd^  —  2d^-\-d. 

6.  /  — 6/+  lly  — 6  and?/«  — 8/+  H>y  —  12. 

7.  14.9*  —  Is^  —  10  and  U*-*  +  245^  +  206'  +  10. 

8.  2;^  +  12 1^  +  22r  +  12  and  r^  +  r^  —  4.7^  -  4r. 

9.  6^«-8-:2-  172; -6  and  12^«  -  5' -  21;^  -  10. 

10.  w^  +  3  m;2  +  4«/;  +  2  and  2  z«;*  +  w^  +  1. 

11.  h^  -  ^2  ^  A  +  3  and  ^*  +  F  -  3^'  -  A  +  2. 

12.  6F+  15F-6^  +  9  and  9 A' +  6F- 51^4-  36. 

To  find  the  lowest  common  multiple  of  three  or  more  integral  poly- 
nomial algebraic  expressions  first  find  the  lowest  common  multiple^ 


REVIEW  249 

Zi,  of  any  two  of  ihem^  then  the  lowest  common  multiple^  X2,  of  this 
result  and  a  third  expression^  and  so  on,  until  all  of  the  given  expres- 
sions have  been  used. 

The  lowest  common  multiple  last  obtained  will  be  the  expression  of 
loivest  degree  which  may  be  divided  without  remainder  by  each  of  the 
given  expressions. 

13.  a^  +  «'  -  lOrt  +  8  ;  «2  -  3rt  +  2  and  ^/»  -  4^/^  ^  5^^  _  ^ 

14.  m^  +  2  am^  +  4  a'^m  +  8  rt* ;  m^  —  2  am^  +  4  a^m  —  8  a^  and 
m^  4-  4  ff "^wi  +  «m^  +  4  rt*. 

15.  2  /r+3  w— 20 ;  6  ^"-25  wH21w+ 10  and  2  w*-5  ^i'+G  7«-15. 

16.  ar^-3^>a;+  2^>^  ic'-5^«!+  4  6'and3a;^-  19  6a;+  28  6'-*. 

Mental  Exercise  XIV.    3 
Review 

Simplify  each  of  the  following  : 

1.  {x  +  y){x'-  -xy  +  y').  3.  (b'  +  c^)  ■-- (b  +  c). 

2.  (a  -  b){a''  +  ab-\-  b^).  4.  (c«  -  cT)  -r-  (c=^  -  ^). 

Distinguish  between 
5.  a  +  6c  and  («  +  6)c.  6.  £c  +  J/-^  +  Wj  {x  +  y)z  +  «^  and 

{x  +  y)(z  +  w). 
Perform  the  following  multiplications  : 

7.  (rtr  +  6  +  x){a  -{-b-x).        9.  {x-y+  \){x  -^  y  -  I). 

8.  (a  +  6  +  7)(a  +  ^^  —  7).      10.  (r«4-6  +  Wi  +  w)(«  +  6— w«  — ^0- 

1 1.  Find  the  continued  product  oix  +  y,  x^  —  y'^,  x^  +  ?/  and  ic*  +  v/*. 

12.  Show  that  x^  —  2  x^  +  aj-  is  the  square  of  a  binomial. 

Are  the  following  expressions  conditionally  or  identically  equal  ? 

13.  3 X  and  2x  +  x.  IQ.  a  +  b  and  b  +  a. 

14.  3ic  and  2  +  1.  17.  a  +  ^  and  b  +  d. 

15.  (a  +  by  and  a*  +  2«6  +  b\     18.  f  and  1. 

Are  the  following  expressions  identical  % 

19.  a«  +  6*  and  (r*  +  6)'.  21.  ^^^  -  ^^'^  and  {a  -  b)\ 

20.  {a^b)\a-by^x^^{a'-Wf.     22.  (oj  -  1)'  and  x"-  -  1. 


250  FIRST  COURSE  IN  ALGEBRA 

Of  the  following  equations,  select  those  which  are  equivalent  to 
the  equation  2  a;  —  3  ^  =  5  : 

23.  4:X  —  Qy=10;Qx  —  9ij=15',8x  +  l2i/=^0;  lOaj— 15?/=25. 

Replace  each  of  the  following  single  equations  by  a  set  of  separate 
equations  which,  taken  together,  are  equivalent  to  it : 

24.  (x  +  6)(x  -  3)  =  0.  27.  (w  +  l)(w  +  9)  =  0. 

25.  (i/  —  2)(i/  +  7)  =  0.  28.  (a  +  G)(a  +  10)(«  +  12)  =  0. 

26.  {z  -  4)(2  -  8)  =  0.  29.  (b  -  2)(0  +  5)(6  -  15)  =  0. 

Construct  single  equations  which  are  equivalent  to  the  following 
pairs  of  separate  equations  : 

30.  x—2  =  0 and «  —  3  =  0.     S3,  w  +  1  =  0  and  W  +  S=0. 

31.  y  — 4  =  0  and  3/  — 6  =  0.     34.  a;  =  1  and  ic  =  2. 

32.  z+  5  =  0 and z—\=0,      35.  y  =  5  and  ?/  =  7. 

Solve  the  following  conditional  equations  for  x,  y,  z  and  w  : 

36.  7a;  — 7  =  0.  38.  3;^— 12  =  0. 

37.  2y  —  8  =  0.  39.  6  -  2  «<;  =  0. 

Show  that  the  following  identities  are  true  : 

40.  2(«  +  6)3(«  -b)  =  G(a^  -  P). 

41.  aia""  -f  b^)  -  a\a  -  b)  =  bicL^  +  b^)  +  b\a  -  b). 


FRACTIONS  251 


•       CHAPTER  XV 

RATIONAL  FRACTIONS 

1.  In  order  to  obtain  as  a  quotient  a  number  previously  defined, 
that  is,  a  positive  or  a  negative  number,  the  dividend  must  be  a 
multiple  of  the  divisor. 

E.  g.    If  the  divisor  be  7,  the  dividend  must  be  some  multiple  of  7,  that  is, 

7,  14,  21,  etc.,  or  -  7,  -  14,  -  21,  etc. 

We  may  obtain  by  actual  diWsion 

21  -f  7  =  3,     35  ^  7  =  5,     _  14  -f  7  =  -  2,  etc., 

the  quotients  in  all  cases  being  numbers  previously  defined,  that  is,  either 
positive  or  negative  whole  numbers. 

2.  From  this  point  of  view  a  combination  of  symbols  has  no 
meaning  when  it  consists  of  a  dividend  which  is  not  a  multiple  of 
the  divisor. 

E.  g.  In  this  sense,  12  -^  7,  3  -^  7,  —  2  -^  7,  have  no  meaning,  since  in 
such  cases  the  division  can  never  be  performed  exactly. 

It  will  be  noted  that  expressions  such  as  those  above  have  the  forms  of 
quotients,  and  by  the  Principle  of  No  Exception  our  idea  of  number  must 
be  extended  to  include  such  quotient  forms  as  numbers.  We  shall  admit 
such  expressions  to  our  calculations  and  reckon  with  them  as  with  ordinary 
quotients. 

3.  Negative  numbers  were  invented  because  of  the  impossibility 
of  subtraction  in  all  cases,  and  broken  numbers  or  fractions  because 
of  the  impossibility  of  division  in  all  cases.  Combining  these  two 
extended  ideas  of  number,  we  are  led  to  the  idea  of  negative  frac- 
tional numbers. 

Thus,  the  idea  of  "fractured  or  broken  numbers"  arises  from 
division  in  a  way  similar  to  that  in  which  the  idea  of  negative 
numbers  arose  in  connection  with  subtraction. 


252  FIRST  COURSE  IN   ALGEBRA 

4.  A  fraction  is  expressed  by  writing  the  dividend  above  the 
divisor  and  separating  the  two  by  a  horizontal  line  or  "  stroke  of 

division."     Thus,  the  fractional  notation  j  and  the  solidus  notation 

ajb  each  express  the  same  as  a  -^  b. 

5.  In  the  quotient  symbol,  or  fraction,  the  divisor  or  part  written 
below  the  horizontal  line  is  called  the  deiioiniiiator  of  the  fraction, 
since  it  mimes  or  danoini nates  the  number  of  parts  into  which  unity 
is  supposed  to  be  divided.  The  number  above  the  line  indicates 
how  many  of  these  parts  are  to  be  taken.  Hence  this  number  is 
called  the  numerator,  since  it  enumerates  or  counts  the  parts. 

6.  When  either  the  numerator  or  the  denominator  of  a  fraction 
is  a  polynomial,  the  horizontal  stroke  of  division  separating  them 
serves  both  as  a  sign  of  division  and  as  a  sign  of  grouping. 

E.  g.  x  +  ^  =  a:  +  (2/  +  z)  ^  3. 

7.  Algebraic  fractions  have  the  same  properties  and  are  governed 
by  the  same  rules  of  calculation  as  arithmetic  fractions. 

E.  g.  As  1/4  is  to  be  regarded  as  being  one  of  the  parts  obtained  by 
separating  or  dividing  unity  into  four  equal  parts,  so  1/6  is  to  be  taken  as 
representing  one  of  the  h  equal  parts  into  which  unity  may  be  divided, 
according  to  our  extended  idea  of  number.  Furthermore,  as  3/4  is  under- 
stood as  meaning  three  of  four  equal  parts  of  unity,  a/b  is  to  be  understood, 
as  meaning  a  of  the  b  equal  parts  of  unity. 

8.  For  a  broken  number  or  fraction,  we  may  write  as  our 

Definition  Formula,  ^  xb  =  a, 
b 

From  this  it  appears  that  the  iwoduct  obtained  by  multiplying 

the  quotient  symbol  by  the  divisor  is  identically  equal  to  the  dividend. 

9.  The  terms  of  a  fraction  are  the  numbers  or  expressions 
separated  by  the  line  or  symbol  of  division. 

10.  Since  zero  can  never  be  used  as  a  divisor,  the  ordinary  laws 
of  reckoning  cannot  be  applied  to  fractions  whose  denominators  are 
zero. 

11.  A  rational  aljjrebraic  fraction  is  the  quotient  obtained  by 
dividing  one  rational  integral  function  by  another. 


FRACTIONS  253 

12.  A  fraction  is  said  to  be  proper  if  the  degree  of  the  numerator 
is  lower  than  that  of  the  denominator  when  the  degrees  of  both  are 
reckoned  in  terms  of  some  common  letter  of  reference. 

2  X 

E.  g.  — — r ,      .,  are  proper  fractions. 

x  -\-  o        cc    -J-  i 

13.  A  fraction  is  said  to  be  improper  if  the  degree  of  the  numer- 
ator is  equal  to  or  greater  than  that  of  the  denominator,  reckoned 
in  terms  of  some  common  letter  of  reference. 

-bi.  g.  — —T ,     j——r    are  improper  fractions, 

but  g  ,    ^  >     , .   ,    ,     are  proper  fractions. 

14.  We  say  that  a  given  value  has  the  form  of  a  fraction  when 
it  can  be  expressed  as  a  quotient. 

E.  g.   Each  of  the  following  fractions  represents  the  value  3: 

6/2,    30/10,    12/4,    3/1. 

Again,  a%/a^,    a*b^/ab^,    7  a*b*/7  a%^^  all  represent  a%. 

15.  To  reduce  a  fraction  is  to  change  its  form  without  altering 
its  value. 

16.  When  a  quotient  can  be  so  transformed  as  to  become  integral, 
the  dividend  is  said  to  be  exactly  divisible  by  the  divisor. 

17.  When  a  quotient  cannot  be  so  transformed  as  to  become 
integral  it  is  said  to  be  fractional^  or  for  emphasis,  essentially 
fractional. 

18.  A  fraction  considered  as  a  whole,  that  is,  as  a  quotient,  must 
be  regarded  as  possessing  quality,  that  is,  as  being  either  positive  or 
negative. 

The  quality  of  a  fraction  as  a  whole  may  be  indicated  by  writing 
either  +  or  —  as  a  quality  sign  directly  before  the  horizontal  stroke 
of  division,  separating  the  numerator  from  the  denominator. 

E.  g.  The  expression  -\ is  to  be  understood  as  meaning  that  the 

X 

fraction  as  a  whole  is  positive,  while  the  expression ^  is  to  be  under- 
stood as  meaning  that  the  quotient  resulting  from  dividing  x  by  2/  —  ^s  is 
negative. 


254  FIRST  COURSE  IN  ALGEBRA 

19.  In  a  fraction,  the  numerator  as  a  whole,  and  also  the  denomi- 
nator as  a  whole,  may  each  be  regarded  as  possessing  quality. 
Hence,  as  each  may  have  a  sign  independent  of  the  sign  of  the 
fraction  as  a  whole,  we  are  led  to  consider  three  signs  in  connec- 
tion with  any  fraction,  as  follows  : 

+  «  -a  .  +a  ■  -a. 

"^4-6'  "^4-6'  "^  -ft'  "^-6' 

_  +  a  _  —  a  _  4-  a  _  —  a 

4-6'  +b'  -6'  -b' 

20.  Since  a  fraction  is  an  indicated  quotient,  the  fundamental 
laws  of  signs  for  multiplication  and  division  may  be  applied  directly. 
(See  Chap.  V.  §§  9,  43.) 

Hence  we  have  the  following  Principles  relatinff  to  the  signs 
of  a  fraction : 

(i.)  The  signs  of  both  numerator  and  denominator  as  wholes  may 
he  changed  from  +  to  —  or  from  —  to  +  without  altering  the  value 
of  the  quotient,  and  hence,  without  affecting  the  sign  before  the  whole 
fraction, 

—  a_      -f- a       4-a_      —a       —  (t  _      -j-  a        4-  a  _      —a 

^•«-  +=r6  =  +  :^'-zi  =  -:^'-ri,-~  +  6'    -b-    +b' 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

The  changes  of  signs  in  the  first  illustration  above  may  be  explained  as 
follows: 

4-^  =  4-[-a]-f[-&] 

=  +  [-a]4-[-6]x(-l)-(-l) 

=  4-[(-«)x(-l)]4-[(-6)x(-l)] 

=  4-  [4-  a]  4-  [4-  &] 

-^  +  b' 
The  changes  of  signs  in  the  remaining  illustrations  may  be  explained  in 
a  similar  way. 

Ex.  1.   Express  ~  ^  ~  ^  as  an  equivalent  fraction  containing  the  least 
^  11  —  m 

possible  number  of  negative  signs. 

Reversing  the  signs  of  both  numerator  and  denominator  as  wholes,  we 

—  X  -  y  _—  {-  X  —  y)  _  X  +  y 


have 


—  {n  —  m)       m  —  n 


FRACTIONS  255 

Since  if  the  sign  of  either  the  whole  numerator  or  of  the  whole 
denominator  be  changed,  the  sign  of  the  fraction  as  a  quotient  will 
be  changed,  it  follows  that  the  sign  of  the  fraction  will  be  restored 
by  reversing  at  the  same  time  the  sign  before  the  fraction  as  a 
whole.     Hence, 

(ii.)  The  sign  before  a  fraction  may  be  changed^  provided  that  the 
sign  of  either  the  whole  numerat&r  or  of  tJie  whole  deiuyminator  be  also 
changed. 

—  a_      +a        +a_      +«        —  a  _      +a        —a_      —a 

The  changes  of  signs  in  these  illustrations  may  be  explained  by  using  a 
method  of  reasoning  similar  to  that  employed  for  the  illustration  of  Prin- 
ciple (i.) 

Ex.  2.    Transform  —  j^ into  an  equivalent  positive  fraction. 

An  equivalent  positive  fraction  may  be  obtained  by  reversing  the  quality 
of  the  fraction  as  a  whole,  and  also  the  quality  of  the  denominator  as  a  whole. 

Accordingly,  we  have  —  t =  — 77 r  = r* 

Exercise  XV.   1 
Express  each  of  the  following  negative  fractions  as  an  equivalent 


13. j 

c  —  b  —  a 


positive  fraction 

1. 

1 

—  a 

2. 

2 

-b 

3. 

—  c 
5 

4. 

1  —a 
2 

5. 

3 
b^b 

fi. 

y-x 

7. 

m 

b-c 

8. 

-5 
a;-4 

9. 

c  —  a 

b  -V  c 

10. 

x-\-y  ^ 
z-y 

1 1 

—  a 

11. 

-6  +  c 

1 0 

1 

14.  - 

15.  - 

16.  - 

17.  - 

18. 


X  —  y  -\-  z 
z  —  X  —  y 

-b  +  a 
—  X  —  y  —  z 

1 
{2/  —  x){x-\-y) 

a 

(a-b)(c-by 

—  a 


b  —  a  +  c  '        {b—a){—b—c) 


256  FIRST  COURSE  IN  ALGEBRA 

Show  that  each  of  the  following  identities  is  true  : 

,9    _J_  = L^.        28.       b-c-a_a-b  +  c 

b  —  a  a  —  b 

20.  ^^,  =T-^'  29. 

c  —  b      b  —  c 

2i.l^  =  -'-Zl..       30.- 


b 

-          b 

b- 

c  ■ 

—  a 

—  (1  — 
-b- 

_a  —  b  +  c 
'  a  +  b  —  c 

c 

_ 

c 

3                   3  '  a(c  —  b)           a(b  —  c) 

^^    V  —  X      X  —  y  „,               1                                1 

22    *- = —-  31.  — 

'  w  —  z~z  —  w  '  (a  —  b)(c  —  b)~      (ii  —  b){b  —  c) 

^_-a-J.      a±b_  32.              1             -             1 


b  —  a      a  —  b  '  {b  —  a){c  —  b)       {a  —  b){b  —  c) 


r..        X                 «  ««  1  1 

24.  = 33. 


—  y        y  —  z  'hi  —  ^f     {^-yy 

2,.--^^-^-^.        34.-        '        -        1 


3      -          3  "•       {y-xf-ix-yf 

26.   ■       ^         = L_.         35.      -'      -        ' 


b  —  a  —  c~-      a-b  +  c  '  (I  —  xY  ~  (x  —  ly 

27.  }  = ^ 36.  1         - 


c-b-a-      a  +  b-c  {b^-ay-      {a^  -  by 

1  1 


37. 
38. 
39. 


{y  —  x){z  —  y)  (z  —  aj)       {x  —  y)(i/  —  z){z  —  x) 
1  _  1 


(a  —  U){c  —  b){c  —  a)  {a  —  b)(b  —  c)(c  —  a) 

a  —  c  _  c  —  a 

(a  -  b)(c  -b)  =  (a-  b)(b  -  c)  * 


^g  (i/  —  ^)Q/  —  ^)  -  (^  —  ?/)(y  —  ^) 


41.  - 


{w  —  z)(w  —  x)       {z  —  w)(w  —  x) 

c  —  a  —  b  a  -^  b  —  c 


{b  —  c  —  a)(a  —  b  —  c)       (a  —  b  +  c){b  +  c  —  a) 


21.  An  expression  is  said  to  be  fractional  if  it  contains  one 
or  more  fractional  terms  ;  if  it  contains  both  fractional  and  integral 
expressions,  it  is  said  to  be  mixed. 


FRACTIONS  257 

An  expression  is  said  to  be  entirely  fractional  if  it  contains 
fractions  only. 

ft  O  V   _L   fit 

E.  g.  T  +  -, is  entirely  fractional; 

°  b      a         z  •' 

while  m^  —  n^  -\ \- -  is  a  mixed  expression. 

22.  Two  fractions  are  said  to  be  equal  or  equivalent  when  the 
terms  of  either  may  be  obtained  from  the  terms  of  the  other  by 
multiplying  or  dividing  both  numerator  and  denominator  by  the 
same  number  or  expression,  zero  excepted. 

E.  g.  I  is  equal  to  ^,  since  the  numerator  and  denominator  of  the  second 
fraction  can  be  obtained  from  the  numerator  and  denominator  of  the  first 
fraction  by  dividing  each  by  4. 

Also,  -^,  =  r»  since  by  dividing  the  numerator  and  denominator  of  the 

first  fraction  by  the   same  expression,  ah,   we  obtain  the  numerator  and 
denominator  of  the  second  fraction. 

Reduction  to  Lowest  Terms. 

23.  A  fraction  whose  terms  are  wholly  rational  and  integral  is 
said  to  be  in  lowest  terms  or  simplified,  when  its  numerator  and 
denominator  have  no  common  integral  factors. 

(The  foUowing  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Consider  the  fraction  rr*  a,  ^,  and  h  being  positive  integers.     Then,  if  a 

and  h  be  prime  to  each  other,  and  also  to  h,  h  will  be  the  H.  C.  F.  of  the 
numerator  ah  and  the  denominator  bh. 

By  the  Laws  of  Association  and  Commutation  for  multiplication  and 
division,  and  the  definition  of  a  quotient,  we  have 

^  =  (a/i)-fW 

=  axh-Tb-^h 
=  a-^bxh-7-h 
^  ah  _a 

hh     h 

Read  backward  and  forward,  this  identity  establishes  the  following 

17 


258  FIRST  COURSE  IN  ALGEBRA 

Principle:  If  both  numerator  and  denominator  of  a  fraction  he 
multiplied  by  or  divided  by  the  same  number^  except  zero,  the  value 
of  the  fraction  ir ill  remain  nnaltered. 

24.  From  this  principle  it  follows  that  a  fraction  may  be  reduced 
to  lowest  terms  if  both  numerator  and  denmninator  be  divided  by  their 
highest  common  factor. 

Ex.  1.    Reduce  18  a%\-  /  Ah  a^bc  to  lowest  terms. 

The  H.  C.  F.  of  iiuinerator  and  denominator  is  found  to  be  Qa^bc. 

18  a^b^c_  (9a^bc)(2b)  Check. 

®°^^  46  a^bc  -  (9  a%c){b  a^)  a  =  2,  fc  =  3,  c  =  4. 

25.  This  operation  of  removing  common  factors  from  both 
numerator  and  denominator  of  a  fraction  by  division  is  called 
cancellation. 

The  student  should  nei^er  strike  out  equal  terms  which  are  found 
in  both  numerator  and  denominator,  unless  they  are  factors  of  both 
the  whole  numerator  and  the  whole  denominator. 

E-ff-    In  — -s — —  we  must  not  strike  out  the  first  terms,  5x^,  nor 
ox-*  —  14 

attempt  to  remove  7  from  21  and  14,  for  neither  5x^  nor  7  is  a  factor  of 
both  the  entire  numerator  and  the  entire  denominator. 
Ex.  2.   Reduce  l2xhf^/(2Ax^y  -  12 xy^)  to  lowest  terms. 
The  terms  of  the  denominator  contain  12  xy  as  a  common  factor.     Hence, 
we  may  write 

12  xV         _         12xV 
24xhj  -  I2xy^~  I2xy(2x-y)' 
Dividing  both  numerator  and  denominator  by  the  common  factor,  12a:y, 


-2x-y  18=18. 

Ex.  3.    Reduce  (6  xhj  -  15  xh/)  /  (10  x^/  -  25  xy^)  to  lowest  terms. 

6 a:«^  -  15 x^y^  _  3 xhf  (2x-5y)    Check.     x=2,  y=l. 
10a;V-25a:?/8  "  5xy^{2x-5y)  |  =  f 

_3ar 
5y 

It  is  sometimes  necessary  to  find  the  H.  C  F.  of  numerator  and 
denominator  by  the  division  process. 


FRACTIONS  259 

Ex.  4.  Reduce  (2x^ -h  x^  -  16x  +  15)/(6 a;»  -  11  z^  +  7 a:  -  6)  to  lowest 
terms. 

By  the  division  process,  the  H.  C.  F.  of  numerator  and  denominator  is 
found  to  be  (2  a;  —  3).     Hence  we  may  write 

2x8+       a:2-16a:  +  15_  (2a;-3)(  a;2  +  2a:-5) 
6a;8-lla;2+    7x-    6  ~  (2a;  -  3)(3x2  -    x  +  2) 

_    a;^  +  2  a:  -  5  Clieck.     x  =  2. 

—  3a;=^—    x  +  2'  i  =  i- 


Exercise  XV.  2 

Simplify  the  following  fractions,  checking  all  results  by  substituting 
such  numerical  values  as  do  not  reduce  the  denominators  to  zero  : 

21   ^. 


1. 

xz 

2. 

ab 

a'' 

3. 

7 
216* 

4. 

20  2; 

5. 

28  a* 
14  a» 

6. 

7. 

25  A* 

5A» 

a. 

^U 

24  c* 

9. 

\^cd 

9«5* 

10. 

ab^c 

11. 

xyz" 

12. 

a'b^c' 
a'b'c' 

13. 

4tac^x 
4.  ex" 

14. 

12X2^ 

4:X1JZ^ 

1  ^ 

Uc'xy 

33a;V^ 

ir. 

19«6*c^ 

lU. 

57  abc^cP 

17. 

a"* 

18. 

b^' 

19. 

^+1 

90 

iK»+* 

22. 

23. 

24. 

25. 

af+y-1 

af'-Y+^ 

26. 

ab 

ac  +  «G? 

27. 

xy 

yz  —  yw 

28. 

m 

rri^  +  m 

29. 

2  a* 

2a  +  2 

Qf\ 

aw  +  a/i 

ic"+®  aic  +  a^ 


260  FIRST  COURSE  IN  ALGEBRA 


ab  +  bc  ■  4a"— 1 

ia  +  Sb    >  (c  +  rf)' 

12c  +  16<i'  c«-rf« 


33.  ^f^^-  39. 
10  a  +  20  s 

34.  «£+i|i.  40. 


35.   .     .    ..o'  41. 


3m  — 9» 

a4-* 

(a  +  by 

x+  1 

w^- 

-w'^ 

(m- 

-n)« 

rf«- 

■1 

(^+1 

«»- 

-  1 

iC»- 

■  1 

a'- 

-b^ 

(^-d* 

7? 

-xiZ-^-f 

x'+y" 

a' 

+  ab  +  b* 

a'-U" 

a^ 

+  2a  +  1 

a^ 

+  '6a  +  2 

b' 

-5^  +  6 

b^ 

—  46  +  4 

c*+6c  +  9 

c' 

+  C-6 

Q? 

-3a;- 10 

43. 

x^-2x+  1 

44. 

y  —  4 

y^-8y+  16 

45. 

a+  1 

a*4-  3a  +  2 

46. 

a;-5 

a;^-  7a;+  10 

47. 

a«-3a^ 

a^- 6a  +  9 

48. 

7W*  +  2  7W* 

m^  +  4  m  +  4 

2a^+  5a  +  3 

5a^+  12a  +  7 

a«+  3a  +  2 

2a"+  5a  +  2 

P-' 

7X:+  12 

36.^— T-  42.^«_^^3 

49.  ^-^-^j--^.  58. 


50.  -    ^-^  ;^.  69. 

^^-         •       ••      •  ^^-    F-5>t  +  6 

52.  '- :  : '  :  :•  ei.  ^._^f,_^^y 

53.  ..       *•    ■    '  62.  ^^_^^^,__^, 

^^-  -^^ ^'  ^^-  (a  +  c)«-6»- 

^        ^^       ^^  „.    (a;-  yy-^^ 

^^'    a;a_  63.^5*  ^4.  ^^  _  (^  +  ^)2 

y-8y+12  a;»  +  ar'+3a;-5 

^^-  y»_92^+ 18*  ^^-       ic'»_4a;+3 

1  +5a;+  Gar'  a;' +  3a;^+8a;  +  2 

1  +6a;+8ar'*  «»  -  2a;'' -  2aj  -  3 


I 


r 


FRACTIONS  261 

26.  If  the  degree  of  the  numerator  be  equal  to  or  higher  than 
that  of  the  denominator,  the  form  of  a  fractional  expression  may  be 
such  as  to  admit  of  transformation  into  an  equivalent  integral  ex- 
pression, or  into  a  mixed  expression. 

E     1  gg  -  ftg  _  (g  4-  bXa  -  b)  Check,     a  =  3,  fc  =  2. 

a  —  b  ~        (a  —  b)  6  =  5. 

=  a  +  6. 

Ex.  2.    Reduce  (a*  +  2  a^ft  +  3  b^/a^  to  lowest  terms. 
By  the  Distributive  Law  for  division,  we  may  find  the  result  by  diyiding 
each  of  the  terms  of  the  dividend  by  the  divisor,  and  write 

g*  4-  2 a^6  +  3 i!>^  _  «»      2a%      36*    Check,    a  =  2,  6  =  3. 
a*  -  g2  "^    gs    "^  a^*  jy  =  y . 

Ex.  3.   Reduce  s— -h — r-5 —    ^  lowest  terms. 

x^  +  ZX  -\-  6 

Using  the  denominator  as  divisor,  and  carrying  out  the  process  for  long 
division,  we  obtain  x  as  an  integral  quotient  and  —  4x  +  2  as  a  remainder. 

^,      .  x*  +  2x^-x  +  2  _  -4x  +  2         Check,     x  =  2. 

^^^*"''  x«  +  2x  +  3      =''"^x2-f2ar  +  3  \\  =  \\. 

_  4x-2 

~^      a:* +  2x4- 3* 

Exercise  XV.    3 

Reduce  the  following  improper  fractions  to  integral  or  mixed  ex- 
pressions, checking  all  results  numerically  : 

1.  '^^-  6.  ^^.  11. 


2-  ,-^-  7-  ,-^-  12. 

3.?!±i.  B.'^-*^-''-       13. 

aj  -f  4 

7i«-5w- 12 
9. 14.       , 

c^-f  9C4-20  c?»4-3<7'g-f3<//+g' 

A  -a  M^4        '  (/^  +  2<^-f^' 


a  -f  3 

0*^1 

b  -1 

a^+1 

a; 

4fi?*4-  2<^-f  1 

2<^ 

A»-3 

'i 

5«» 

a 

-b 

z' 

-!£;* 

z 

—  ^ 

8' 

+  ^ 

8 

+  ^ 

«* 

262  FIRST  COURSE  IN  ALGEBRA 

Reduction  of  Fractions  to  Equivalent  Fractions 
having  a  Common  Denominator 

27.  Two  or  more  fractions  are  said  to  have  a  common  denomi- 
nator when  their  denominators  do  not  differ. 

T^                           X       .  z              a  .     c  +  d 

E.g.  -  and  -,  or  -5 jj,  and  -^ — rx. 

28.  The  lowest  common  denominator  (or  L.  C.  D.)  of  two 

or  more  fractions  is  the  lowest  common  multiple  (or  L.  C.  M.)  of 
their  denominators,  used  as  a  new  denominator. 

29.  Any  number  of  separate  rational  fractions  may  be  transformed 
into  equivalent  fractions  which  have  equal  denominators. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  a/h  and  c/d  represent  two  fractions,  and,  for  the  present,  let  their 
terms  a,  6,  c,  and  d  be  rational  and  integral.  Let  the  L.  C.  M.  of  the  de- 
nominators b  and  d  of  the  fractions  be  Z,  and  let  bm  =  L,  and  dn  =  L. 

In  these  identities  the  letters  m  and  n  represent  integers  since  the  lowest 
common  multiple  of  two  integers  is  an  integer,  and  m  and  n  must  be  prime 
to  each  other,  for  if  they  were  not  L  would  not  be  the  lowest  common 
multiple. 

Multiplying  both  numerator  and  denominator  of  the  first  fraction  a/bhy 
the  factor  m,  necessary  to  produce  the  L.  C.  D.  of  the  denominators  b  and  d, 

we  have  a  _  am  _  am 

b  ~  bm  ~  L 


Similarly, 


c  _cn  _  en 
d  ~  dn~~  L 


The  two  original  fractions,  a/b  and  c/d,  are  now  expressed  as  the 
equivalent  fractions,  am  j  L  and  cn/L  respectively,^  which  have  equal 
denominators,  L. 

From  the  proof  above  we  have  the  following  : 

To  reduee  two  or  more  fractions  to  equivalent  fractions  having  the 
lowest  common  denominator,  first  find  the  lowest  common  multiple  of 
their  denominators.  Then  multiply  both  the  denominator  and  numer- 
ator of  each  fraction  by  the  number  or  factor  which,  when  taken  with 
the  denominator  of  the  fraction,  will  produce  as  a])roduct  the  required 
lowest  common  multiple  of  all  of  the  denominators. 


FRACTIONS  263 

Ex.  1.    Reduce  2/3(1,  4/5a^,  6/ 15 a^  to  equivalent  fractions  having  the 
L.  C.  D. 

The  L,  C.  D.  is  15  a'.     Hence  we  have 


2 
3a 

= 

2. 
3a 

5a2 
•5a2 

= 

10  a2 
15  a3 

4 
5a2 

= 

4- 
5a2 

3  a 
•3a 

= 

12a 

15  a' 

6 
5a8 

= 

6 

15  a« 

Let  the  student  check  the  example  above. 

Ex.  2.  Reduce  the  fractions  (a  —  b)  /  (a  +  b),  (n  +  b)  /  (a  —  b)  to  equiva- 
lent fractious  having  the  L.  C.  D. 

The  L.  C.  D.  of  the  two  fractions  is  found  to  be  (a  -f  d)(a  -^  b). 
Therefore: 

a-b_(a-b)(a-b)  _  (a  -  b)^      Check.   a  =  3,  6  =  2. 

a-i-b"  (a  +  6)(a  -  b)  -  a^  -  b'^  '  ^  =  f 

qjmnavW  ^  +  ^  -  («  +  b)(a  +  b)  _  (a +  6)2     Check,    a  =  3,  6  =  2. 
aimiiaiiy,  ^  _  ^  =  ^^  _  ^^^^  +  ^j  =  «2  _  ^2  '  5  ^  5. 

The  student  will  find  it  a  good  plan,  when  transforming  a  fraction 
to  an  equivalent  fraction  having  a  different  denominator,  first  to 
copy  the  fraction  in  its  original  form  and  then  to  make  the  necessary 
alterations  by  mserting  the  proper  factors  in  both  denominator  and 
numerator.  This  is  better  than  to  attempt  to  copy  the  firaction 
and  make  the  transformation  at  the  same  time. 

E.  g.    Thus,  in  the  example  above,  we  may  write  as  a  first  step, 

a  -f  ft  _  (a  +  6) 
a  —  6  ~  («  —  b) 

As  a  second  step  we  will  insert  the  factor  (a  +  b)  in  both  denominator 
and  numerator,  as  below: 

a  +  h  _  (g  +  b){a  +  b)  _  (a  +  by 
fc^b-{a-  b)(a  -\-b)-  d^  -  6^  * 


13. 

a    b 

14. 

y  X 

15. 

be'  cd 

16. 

yz   zw 

17. 

1        2 

Sab'  4tbc 

264  FIRST  COURSE  IN  ALGEBRA 

Exercise  XV.    4 

Reduce  the  fractions  in  each  of  the  following  groups  to  equiva- 
lent fractions  having  the  L.  C.  D.  : 

a    a  -56 

^    b    h  2     7 

5    7  a^    a 

o    c     c  q    3    4 

6.  —  J  —  •  y.  — »  —  • 

8    12  X   y 

4.  -r '  7:7:*  1^'  ~'  — 
5    20  z    w 

5.  ->  7"  11*  TT"'  ir~ 
a    6  2a    7a 

.22  19     ^     1<^ 

6.  -^»  —^'  12.  — r>  "T- 
«''   y'^  ab    be 

a    6    c 

XXX 

'  2y'  4^^'  6y 

b     b^    b^ 
21.  — >  — > 


««      1       1       1 
22.  — I  — >  — 

2£c    3y    4:Z  x  —  y    x  +  y 

^o    ^   y   ^  on  _i 1 

2/    2;    a;  ix—y){y  —  z)    {y—z){z—x) 

24.  -,  A,_l.  31.        2  3  4 

'  be    ca    ab  *ar^  —  9£c  +  3'ic— 3 

25.?,J-,A.  32.       1  1  1 


a        «« 

56c    lOc^ 

26. 

X 

y-Vz 

Z  +  X 

27. 

1 

x^\' 

1 
V-1 

28. 

1 

1 

a'-b^ 

99. 

a;  + ?/ 

X  —  y 

a    2a   a^  a  —  b    b  —  c   c  —  a 


FRACTIONS  265 

30.*  If  all  of  the  terms  of  two  equal  fractions^  xjy  and  n/d^  he 
positive  whole  number  Sy  then  if  njdhe  in  lowest  terms,  it  follows  that 
x=^  pn  and  y  =  pdy  where  p  represents  a  positive  whole  number. 

For,  if       1  =  1  (1) 

then  ^  =  ^-  (2) 

Or,  the  quotient  obtained  by  dividing  ny  by  c?  is  a  positive  whole  num- 
ber a;,  and  since  n  is  prime  to  d,  y  must  be  some  multiple  of  d,  say  y  =  pd, 
where  p  represents  a  positive  integer. 

Hence,  (2)  becomes  x  =  -~,   ov  x  =  pn. 

31.*  /w  particular y  if  ^ly  is  in  lowest  terms,  x  and  y  can  him 
no  common  factor  p ',  hence,  putting  p  =  1,  we  have 

x  =  n,  y  =  d. 

Addition  and  Subtraction  of  Fractions 

32.  Applying  the  Distributive  Law  for  Division  (See  Chapter  V. 
§§  58,59),  a,  b,  and  d  being  taken  as  positive  integers  as  in  other 
proofs,  we  may  write 

-j-\--j  =  a-r-d+b-7-d      and      -:  — ,  =  a  -r-  d  —  b  -t-  d 
d      d  d      d 


=  {a  +  b)^d 
^a  +  h 
-      d 


=  (a-b)-i-d 
_a  —  b 


Hence,  the  sum  of  two  fractions  having  a  common  denominator  is 
a  fraction  having  for  numerator  the  sum  of  the  numerators  of  the 
given  fractions,  and  for  denominator  their  common  denominator. 

Also,  the  difference  of  two  fractions  having  a  common  denomina- 
tor is  a  fraction  having  for  numerator  the  difference  of  the  numera- 
tors of  the  given  fractions,  and  for  denominator  their  common 
denominator. 

*  Thi9  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time, 


266  FIRST  COURSE  IN  ALGEBRA 

Ex.  1.   Find  the  sum  of  a /be  and  hfcd. 
The  lowest  common  denominator  is  bed. 


tor  is  bed. 

Check. 

a  =  2,  6  =  3, 

a       b    _ad       b^ 
bc'^  cd-  bed  "^  bed 

c  =  4,  rf  =  5. 

Hence, 

Writing  the  sum  of  the  numerators  _fid  +  b^ 

over  the  lowest  common  denominator, "~     bed 

2a-\-b      a-2b  _(2a  +  b)4  +  (a-2b)3 

^''•^'      •  ~'3b"^~4b~=  12b 

_    „       .        ,          ,  .  ,.      .                       8a  +  46  +  3a-66 
Perforimng  the  multiplications,  = ^oh 

_,..,.,                                              11a -26       Check,    a  =  3,  6  =  2. 
Combining  like  terms,  =  — —r i4  —  ii 

Instead  of  finding  a  common  denominator  at  once  for  all  of  the 
fractions  of  a  given  set,  it  is  sometimes  desirable  to  combine  the 
fractions  by  groups,  and  after  reduction  of  the  groups  separately,  to 
combine  the  results  thus  obtained. 

^  «.      ,..  3  6  5  3 

Ex.  3.    S.mpl.fy    ^^  +  ^-f  -  ^-j-j  -  ^-^  • 

The  given  fractions  may  be  rearranged  as  follows : 
3  3  5  5_3(x-2)  3  (x  +  2) 


x  +  2     a;-2^a;-l      x  +  l  "  (x  +  2)(a:  -  2)       (a:  -  2)(a:  +  2) 

5  (x  +  1)  5  (x  -  1) 

■^  (x  -  iXx  +  1)       (x-\-  iXx  -  1) 
_3(a;_2)-3(a;  +  2)      5  (a;+ 1) -5  (x- 1) 
-      (a:  +  2)(x-2)      "^       (a:  +  l)(aj  -  1) 
3a;- 6 -3a; 


Ex.  4.   Simplify -, j- 

^     ^      a  +  b  2a  — b 

The  L.  C.  D.  of  the  fractions  is  (a  +  6)  (2  a  -  6). 


x^-4 

x2-l 

_    -12            10 

-  a:2  ^  4  '  a;2  -  1 

-  12  (a:2  _  1)  +  10  (a;2  - 

-4) 

(x2-4)(x2-l) 

-  2  x2  -  28 

Check.     X  =  3. 

-  (x^  -  4)(x2  _  1) 

-M  =  -M- 

2x2  +  28 

-       (a;2_4)(x2_l) 

.      3a-46 

FRACTIONS  267 

Hence, 

2a-Sb  _  3a-4b  _  (2  a  -  Sb)(2a  ^  b)  _  (3  a  -  4  h)(a  +  b) 
a  +  6         2a -b  ~     (a  +  fe)(2a-6)     ~   (2  a  -  b)  {a -\- b) 

_  (2  g  -  3  6) (2  g  -  6)  -  (3  fl  -  4  &)(a  +  &) 
""  (g  +  6)(2a-6) 

_  (4a^  -  8ab  +  36'^  -  (3g'-^  -  ab  -  4b^) 
~  (g  +  6)(2g-6) 

_.  4a3  -  8a&  +  35'^  -  3  g'^  +  q&  +  4  6^ 
•"  (g  +  b)(2  a-b) 

__  a^-7ab  +  7b^  Check,    g  =  4,  6  =  2. 

-(g  +  6)(2a-6)*  -  i  =  -  f 

Exercise  XV.     5 

Simplify  the  following  expressions,  reducing  all  results  to  lowest 
terms,  and  check  numerically  : 


i-r  +  r 

b      b 

10.^  +  1  +  ^. 
yz      zx      xy 

b       c 

11.   '^       *       *' 

a-b      a  +  6      d^-b"" 

3.^  +  *-'. 

12-      \          \- 

c       a 

a  —  b      a  +  b 

4.^  + A. 

be      ca 

13-  i/+  !  • 

oca 

a  +  b      a  —  b 

g    «  +  ^  1  b  +  c 
c             a 

^^-      2       '       3     • 

„   a  —  b      b  —  c  ■ 
'■      a     -     b     ■ 

^^    ,_3,^,  +  y_ 

a  —  b      b  —  c 
*•      a             b     ~ 

c  —  a 

a             6 

c 

9."-*. 

^  y  ^  3,_ 

xy      yz 

5              3 

268  FIRST  COURSE   IN  ALGEBRA 

a  +  2      a-^^  29    ^^  +  ^J^- 

2                 2  on    ^  "^  ^      c  —  d 

^^'  ^MHl  ~'^f^^'  ^'  7^^  "  7+d' 

4              5  ^^    a;  +  1,  ar*  +  1 

ic  —  4      a;  +  5  x—  1      ar— 1 

a;+2      x+3  '       a+1       "^       a-1 

23.-^ U'  33.       ^-1                ^  +  " 


-1       w  +  l  «=*  —  «+!       0"  +  a+1 


oA    a^  +  y  ,  g-y      g  +  y      „^       5,8  3y 

^^•"6""^      3  9     *     "^^'x  +  y^x-y      x"  -  f 

25   ttl ^  +  _^.  35  3  6 9_ 

^^'  b-y      b  +  y^y^-b^  x{x-\-3y  x(x-'6)     x'-'d 

26.-l,-,^i-,.  36.-A,+      ^  ^^^ 


a^  +  aZ>      6^*  +  a6  a;  +  5      a;  -  5      a;^  -  25 


3a! -ar"      ar"  -  9  a  +  6      (a +  6)'      {a  +  b)' 


a  — 6      a  +  b  '  (a  +  xy     (a  —  xY     x^  —  a^ 

a^ 


39. 


(c  —  a)(c  —  6)       (a  —  c)(6  —  a)       (6  —  c)(a  —  h) 


X  y  z 

(x  —  y){x-z)  (i^  —  z)(t/  —  x)  {z  —  x)(z  —  y) 

b+1  ,  c4-  1  a  +  1 

41.     -^ 7T7 T  + 


42. 


(b-cXb-a)       (c-bXc-a)       {a  -  b){c  -  a) 

1 + 1 + 1 , 

(b  —  c){b  —  a)       (b  —  c)(c  —  a)       (b  —  a)(c  —  a) 


4  2  2 

43.    7 7^ r^  + 


(a  —  l)(a  —  3)       (a  —  2)(3  -a)       (2  -  a)(l  -  a) 


44.    ^ "^ ^-  +  -^ 

a4-4      a  —  5      a  —  4      a  +  5 

45.^+      '  '  ' 


FRACTIONS  269 


x  —  2      X— 1       x+  2      x+  1 

Reduction  of  Integral  or  Mixed  Expressions  to 
Fractional  Forms 

33.  Any  integer  or  an  integral  expression  may  be  written  in  the 
form  of  a  fraction  having  any  required  denominator. 

This  is  because  successive  multiplications  and  divisions  by  the 
same  number  or  expression  produce  no  change  in  the  value  of  a 
given  number  or  expression. 

In  symbols,  a  =  axb-^b  =  ~» 

Hence  a  given  number  or  expression  may  he  expressed  as  a  fraction 
having  a  required  denominator  provided  that  the  numerator  of  the 
desired  fraction  is  the  product  obtained  by  multiplying  the  given 
number  or  expression  by  the  required  denominator. 

2  6     h^ 
Ex.  1.    Reduce  1  H h  — ,  to  an  improper  fraction. 

26 
Expressing  1  and  —  as  fractions  having  for  denominators  the  L.  C.  D. 

of  a  and  a^,  which  is  a^,  we  have 

,       26      62      a^      2a6      6''  ^.u    i       t  4.         1.0 

1+  —  +  ^  =  -^  +  — r  +  -2  Check.     Let  a  =  6  =  2. 

a       a^      a*        a^       a^ 

_  gg  4-  2  a6  +  6^*  4  =  4. 

Exercise  XV.    6 

"Write  the  following  expressions  as  fractions  with  the  denominators 
indicated  : 

1.  3  with  denominator  a.  5.  xy  with  denominator  xy. 

2.  5  "  "  10  a.       6.        m  +  n        "        "       m  +  n. 

3.  7  "  "         x-\-y.     7.  x+y        "        "       x  —  y. 

4.  ah        "  "  a.  8.  a''-\-ab+b^        "        "        a-  b. 
Reduce  the  following  mixed  expressions  to  the  forms  of  improper 

fractions,  checking  results  numerically  : 


270  FIRST  COURSE  IN  ALGEBRA 


9. 

..|. 

18. 

"=+^+4/ 

10. 

b 

a • 

c 

19. 

«  +  *  +  a  +  6- 

11. 

a  +  2  +  i- 
a 

20. 

h   W   +    »• 

12. 

a:-2  +  -- 

X 

21. 

al,-^^      - 

13. 

a 

22. 

.^4^^      ^• 

14. 

m 

23. 

.'2  +  ^-^^- 

15. 

24. 

,       «^  +  ^^ 
a   +  6 

16. 

2               «!«" 

25. 

,       2a  +  3 
tt  —4 

7W  —  w 

17. 

,    .   6iB+  3 
a  — 4  +  — 

26. 

r^  +  r  +  1  +  ^-r 

bx 

34.  The  following  principles  are  fundamental  to  the  processes  of 
multiplication  and  division  involving  fractions. 

Principle  I.  To  multiply  a  fraction  by  a  whole  number y  ive  may 
either  multiiyly  the  numerator  alone  by  the  given  multiplier^  or  divide 
ths  denominator  alone  by  the  given  multiplier. 

That  is,  «xc  =  5^^ 

or,  ~Xc  ~ 


fr  '    -  -  ft  _f.  c 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 
For  convenience   of  proof  we   shall   assume   that   the   numerator   and 
denominator  of  the  fraction  a/b  and  the  multiplier  c  represent  positive 
integers. 

In  the  proof  below,  which  depends  upon  the  Fundamental  Laws  for  Mul- 
tiplication and  Division  and  also  upon  the  definition  of  a  fraction,  the 
required  product  is  obtained  at  the  left  by  multiplying  the  numerator  alone 


FRACTIONS 


271 


by  the  given  multiplier,  and  at  the  right  by  dividing  the  denominator  alone 
by  the  given  multiplier. 


We  have  ■rxc=(a-^h)xc 
=  a  X  c  -^  b 
=  (a  X  c)  -r  ^ 
axe 


and  also 


X  c 


(a-^h)  X  c 

=  a  -7-  fe  X  c 
=  a  ^  (&  -f  c) 
a 


b^c 


E.  g.   Multiply  5/24  by  3. 

The  product  may  be  obtained  in  either  of  the  two  following  ways  : 


Multiplying  the  numerator  alone 
by  the  multiplier,  we  have 


Dividing  the   denominator    alone 
by  the  multiplier,  we  have 

Ax3  =  2ji3  =  f- 


Principle  II.  To  divide  a  fraction  by  a  whole  number  we  may 
either  divide  the  numerator  alone  by  the  given  divisor ^  or  multiply  the 
denominator  alons  by  the  given  divisor. 

That  is, 

or, 


a 

c 

= 

a 

-r 

c 

h  ' 

b 

9 

a 

c 

= 

a 

h 

h 

X 

c 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

In  the  following  proof,  where  a,  i,  and  c  represent  positive  integers, 
the  quotient  is  obtained  at  the  left  by  dividing  the  numerator  alone  by  the 
divisor,  and  at  the  right  by  multiplying  the  denominator  alone  by  the 
divisor. 


We  have  T^c  =  (a-j-ft)-rC 


and  also 


=  a  -f  ft  -f  c 
=  a  -^  c  -^h 
=  (a  -f  c)  ^  6 

-     h 
E.g.   Divide  15/17  by  5. 
Dividing  the  numerator  alone 
by  the  divisor, 
15^5 


-I-  c  =  (a  -^  ft)  -f  c 
=  a  -^  (h  X  c) 


we  have    ^f  ^  5 


17 


=  iV 


b  X  c 

Multiplying  the  denominator 
alone  by  the  divisor, 
15 


we  have  |^  -^  5  = 


17x5 


=  il  =  lV 


272  FIRST  COURSE   IN  ALGEBRA 

Multiplication  of  an  IiiteKfer  by  a  Fractional  Miiltiplior 

35.  To  multiply  one  number  by  another  is,  by  the  extended  defi- 
nition of  multipliwition,  to  perform  upon  the  multiplicand  exactly 
those  operations  whioh  must  be  performed  upon  the  unit  of  positive 
numbers  to  produce  the  multiplier. 

K.  g.    Let  it  1)0  rwjuired  to  multiply  a  by  c/iL 

To  obUiiii  the  imiltiplier  from  +1-,  we  may  first  divide  +1  into  il  eipial 
partA,  tittch  iHiuul  to  ^\/d^  and  then  take  c  of  these  parts  as  suunnanda,  juul 
thus  obUiiii  liu'  multiplier  c/d, 

Hruie  to  multiply  a  by  r/r/,  wo  may  first  divide  the  multiplicand,  a,  into 
d  ei^ual  i)art8f  each  iMirt  being  represented  by  a/d,  and  then  take  c  of  these 

parts  08  summauds.    The  result  thus  obtained  is  the  desired  product      -  -. 

Thatis,  ""^'d-J^ 

Uenct\  to  multiply  a  whole  number  hff  a  fraction,  mnUJphi 
the  whoU'  nnndx'r  hij  the  nunu' rutin'  of  the  fraction  and  divide  the 
reanft  by  the  denominatm', 

E.g.   Multiply  12  by  2/3. 

12  X  2 
We  have  12  x  |  =  ^-^  =  V  =  »• 

Multiplication  of  One  Fraction  l)y  Another 

36-  I-^'t  it  be  ritpiiixid  to  multiply  ajb  by  c/t/,  regarding  «,  A,  o,  and  (/as 
jwsitive  whole  numlx^rs  in  order  to  simplify  prt)ofs. 

Applying  PrincipU»8  I.  and  II.,  §  lU,  and  applying  a  course  of  reasoning 
similfU'  to  that  uswl  above,  it  appears  that  we  may  perform  the  operation  l)y 
tirst  se^wmtiug  a/b  into  d  e<puil  parts,  each  of  these  parts  being  ix^presented 
by  a  /(/>  X  (/)• 

\\y  Trinciple  II.,  §  34,  taking  e  of  these  ports  as  summands,  we  obtain  as  a 
final  n>suU 

a/hd  +  a/hd  + tor  summands,  that  is,  ac/hd. 

Hence,  we  may  wnte         i:  x  :3  S  x — -• 
''  o     a     o  X  a 

Accordinglt/y  tlie  product  of  two  fractions  h  a  fraction  whose 

numenrtor  in  the  pn>dfirt  of  the  tjiveu  numerators,  and  whose  denomi- 
nator is  the  /o-odttef  (f  the  (jir,  n  dt  luoiii mttors. 

37.  It  may  be  shown  that  fractions  obey  the  Commutative  Law 
for  Multiplication. 


FRACTIONS  273 

Ilencey  to  find  ilie  product  of  two  fractiom^  eitfier  fraction  may  ha 
used  as  multiplier. 

„     J  ba^b      2ahx  _(r,aVj)('2ahj:)  Check. 

_  h)aV/^x  x=y=i  I. 

*•  ^'     c^  _  d'^      ^  io  (,^2  _  l,-^)  -  10  («  4:  ,l,)(c  -  «:/)(^/  +  /y)(a  --b) 

CancMjIliiipj  factoPH  w>rinnon  to  ])(tiU  _(«-♦-  ft)(«  —  ^/)    Check*  «=  ft,  &  =  2, 
numerator  and  denoiriinator  "  2(a  —  />)(c  -f  ^/)  '  c  =r  4,  ^/  =  3, 

1  =  1- 

38.  Any  factorH  which  occur  in  the  uumer&tor  of  one  fniction 
and  in  the  denomiimt<jr  of  another  may  be  cancel le^l  at  the  oiitnet, 
instead  of  being  carried  over  into  the  terms  of  the  re^juirerl  product 
and  then  stricken  out. 

E,  g.  In  Ejatrnplfi  2  of  §  37,  5  in  the  niunerat^ir  of  the  fIrHt  fnuition  may 
be  caiicellerl  with  5  which  i«  a  fty^tor  of  the  denominator  of  the  w;eond ; 
(a  +  h)  in  the  numerator  of  tiie  lifHt  fnw;tion  with  (a  -f-  h)  in  the  da- 
nominator  of  the  second;  and  (c  —  d)  in  the  denominator  of  the  firnt 
iiaction  with  (e  —  d)  in  the  numerator  of  the  second  fraction. 

Mental  Exercise  XV.    7 

ExpreM  each  of  the  following  products  as  a  single  fraction  in 
lowest  terms : 


1.  'i  X  d. 

C 

6..nxl- 

11.   IH./..x(-±) 

2-  p  X  y- 

7.  2X^i. 
0 

^^•^^K^.o^' 

3.  ?;  X  Zc. 

h 

8.  4X-- 

-^l• 

4.  -^  X  2»*. 
10  « 

9.  tf  *  X      ' 

/y       4 

5.  ^  X  c^d. 
c 

10.  16iB*X 

-y 

Hx 

-Jxf.- 

Vti 


/4 

FIRST  COURSE  IN 

ALGEBRA 

16. 

n^  d 

26.4xi. 
ao       ac 

17. 

d      c 

22.  -X^- 
X       be 

27.   4-  X  — .■ 

18. 

y      z 

23.  —  X  — ^. 

«-i<-iy 

19. 

a       h 

24    2«       4c 

«-PH) 

20. 

c  ^  0' 

25.  ±  X  ^^ 

«2/       z 

»-ix?- 

-{-m 

^y'\ 

35. 

a      b      c 
b      c      a 

-  (f  )(- 

acy\ 
Ox)' 

36. 

5^  y  ^2 

-  l-iX 

aczj 

37. 

2a       5^6.^. 
3        4  c        e 

"•Mx| 

38. 

a       ac      ab 
r-  X  -J-  X  — 
be       b        c 

Exercise  XV.     8 

Express  the  following  products  as  single  fractions  in  lowest  terms, 
checking  all  results  numerically  : 

6a'b       25ar'y  5xz'Syz'         7  z* 

-I. 5    /N    rrZ 7"  *  t).     -— S    X    ~T. TT    ^ 


bary^       IHab  '  21/       8  a;'       30a;y 

^    5a'//       14^^  „    a^-b  6 

2.  ^^-^  X  T^^  •  7.  — ^  X 


lo'if       Iba^b  2  d'-b 

^    12abc  ^^  9a^yz  S.  ^  ""  ^'  x      ^ 

7  y2;2<7       28  ^6^  '       10           iK  ~  3 

13  rt^^       6r'«7W  «        35              ^ 

4.  -— ^r-  X  — -717—  •  9.  — —  X 


6  rs'^m      1 1  kHiP  '  x -Y  y      x  —  y 

^c^m?        2^>'         2cc?  a  +  b      a  —  b 

4  6cc?       3  acx      3  ate  '  c  +  c?      e  —  d 


FRACTIONS  275 

14         cc  +  1  x^  —  xy  +  y      ao  +  b 

12.  ^X^.  22.»-+^-X^^ 

iC^  —  1  17  £C  —  ?/         X'^  +  £C<2 

13.±±ix^.  23.%±fx^^ 
ar— 1      rt+1  a^  —  b^       a  -\-  b 

^^    a^+2a+l^     Q>xy  n^^-j^      x{x  ■\- y) 

14. X   —5  -*  i54.  ;  X  • 

3iB3^  a  —  1  ^  -\-  y  m  —  n 


2 «  +  1       a?*^  —  9  ?^^  *      a6ca;         8  a;  +  3  3 


4a  ~  6       m'^  —  25  72'^  a;  +  5       5  o^  —  15 

^^'  ^"=T^  ^   16a'^  -  6^  *  «  -  a       10  +  2a5' 

a;'+  7a;+  12  ^    ^^^  2a^-M^        a=^  -  ^^^ 

17.  i  X       ■    — •  zi'      ,0    .       1     X 


a6c  a; +3  />' +  a^        a'  — 4a;2 

(«  +  2)^  «  +  5  m^  —  2  mw  +  w^  ?w^ 

1^'       9    i     Ti         i     7T   '^  i     IT*        '"^'  0    !  '^       Q  5 

a^  +  8  a  4-  1 5      <f  +  2  m'^  +  7?27i  m^  —  n^ 

a"  ■\-  ?>a        Sabc  a;^  +  a;  -  6       a;'  +  a;  -  12 

"Sbc  a  +  3  '  x^  —  X—  G       x^  —  x—  12 

a^  +  ab  +  b'       (x  +  yy      ,.       (a-b)b         xy(a' ^  b") 
^^'         x+y  a'-b^'  a^-2ab+b^     a^  +  2ab  +  b^' 

(a-b)b  a(b  +  a)  xz(a' -  b') 

^^-    a^^b^-\-2ab^  a^-\-  b"-  -  2ab  a%^y 

g'^  —  4  (^  +  3       g'^  — 4(^  +  4      a'  —  2a  —  ^ 
(^'^  +  ^^  +  2aZ>-c^      a^-2ac-^>'  +  c2 


^2  _  2^c  -  ^2  -  c^       ^^'  -  2^c  -  a'  +  c^ 

Power  of  a  Fraction 

39.   A  fraction  may  be  raised  to  a  power  by  applying  the  defini- 
tion of  a  power  and  the  principle  for  the  multiplication  of  fractions. 


276  FIRST  COURSE  IN  ALGEBRA 

—  \     =  — ,  771  being  considered,  for  the  'present,  a  positive  integer. 

/3  a%'k\^_  38  (a8)8(&'^)M  Check. 

•  \JW)  -        2*  {d^)^  az=h  =  c  =  d  =  2. 

__  27  aV)h*  1728  =  1728. 

-      8iP 

Since  the  converse  of  any  identity  is-trne,  it  follows  from 
\h/        6'»  b'»      \bj 

^'  2-  (x  +  3)-^        =  [      x  +  S      ) 

_r(x  +  3)(a:  +  2)1'^     Check.     Let  a:  =  2. 
"L       (^  +  3)        J  16=16. 

=  (x+  2)2. 

^     „     X,  a^- 10a +  25      ^,  r    r     ^' 

Ex.  3.   Express  -^ r 77;  as  the  power  of  a  fraction. 

^        a^+    8a  +  16 

aa-lOa  +  25  _ (a  -  5)^  Check.     Let  a  =  6. 

a«+    8a +  16 -(a +  4)2  ik  =  liir- 

Mental  Exercise  XV.     9 
Express  the  following  powers  of  quotients  as  quotients  of  powers  : 

1. 


■  (1)' 

'■  (-!)•■ 

-  (!)'• 

'•  (ly- 

-(O- 

"■  (I)' 

■  (-ly- 

'-ay- 

"■  (?)■• 

(-0' 

■«  (Sf- 

-  {-'if- 

-Gy- 

■■(-ly- 

"i-ij 

■■  (O- 

-(-O- 

.» (ny- 

FRACTIONS  277 

--ay-  -(^J-  -©' 

Express  the  following  quotients  of  powers  as  powers  of  fractions  : 

25-  ^-  34.  §-■  43.  ^,. 

25  625  mV 

26.  tt;*  35.  -— -•  44.  — s-- 

49  400  a" 

2^-  81'  ^^-  ^"  *^-  8^- 

28    -A.  37   111.  Aa    81 '^y. 

^*'-       125  .   ^^-  256 

^^-  81  ^^-324 

30.  -JL.  39.  -^2« 


*u. 

16  a^^^ 

47. 

8a«6« 
27  a;y 

48. 

a=^  +  2«6  +  ^>' 

(«^  +  3^)^    • 

AQ 

(m  -  nf 

c'  +  ^cd+d^ 

50. 

{a  +  hf 

(a'  -  by 

K1 

(x-yY 

32  1000 

27  4 

qo    144  ,,    16  k" 

''''•169'      .  *^-  "25""  "^-  i^-ff 

^.  If  n  is  prime  to  d^  then  the  fraction  n  /d  will  he  in  lowest 
terms  and  every  positive  integral  power  of  the  fraction  n/d  will  be  a 
fraction  in  lowest  terms. 

Consider         |  —  I  =  — ,  in  which  n,  p,  and  d  are  positive  integers. 
\d/  -OP 

Since  n  and  d  have  no  factors  in  common,  the  powers  ^and  r^P  cannot 

have  any  factors  in  common,  and  hence  nP /  dP  must  be  in  lowest  terms. 

Division  of  a  Whole  Number  by  a  Fraction 

41.  To  divide  one  number  by  another  is,  by  the  extended  defi- 
nition  of  division,  to  perform  upon  the  dividend  exactly  those 


278  FIRST  COURSE  IN  ALGEBRA 

operations  which  must  be  performed  upon  the  divisor  to  produce 
positive  unity  ('''1). 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  it  be  reijuired  to  divide  a  by  c/rf,  a,  c,  and  d  being  taken  for  conven- 
ience as  positive  whole  numbers. 

To  obtain  +  1  from  the  divisor  c/d  we  may  reverse  tlie  steps  which  would 
have  to  be  taken  to  obtain  c/d  from  + 1,  that  is,  we  may,  by  applying  Prin- 
ciple II  ,  §  34,  divide  the  fraction  c/d  by  c,  obtaining  as  a  quotient  1  /  d  as 
follows : 

c  .     _  c  -f  c  _  1_ 
d  '      ~      d      ~  d 

Multiplying  this  result  by  d,  we  have  by  Principle  I.,  §  34, 

Hence,  to  divide  a  by  c/d  we  may  perform  upon  it  successively  the  oper- 
ations above  •  that  is,  first  dividing  a  by  c,  we  have 

a 
a  4-  r  =  -> 
c 

and  then  multiplying  this  result  by  (/, 

,  a       J      ad 

we  have  -  x  d  = 

c  c 

,,  c      ad 

Hence,  a  -j-     =  — • 

d       c 

That  is,  to  divide  a  whole  nutnher  hy  a  fraction,  we  may 

multiply  the  wlwle  number  by  the  denominator  of  the  fraction  and 
divide  the  result  by  the  numerato?'.     (Compare  with  §  35.) 

Division  of  One  Fraction  by  Another 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time. ) 

42.  Let  the  terms  of  two  given  fractions  be  represented  by  the  positive 
integers  a,  b,  c  and  d.  ' 

Representing  the  fractions  by  a/6  and  c  j d  we  may  write 

CL         C 

=  (a  -^  b) -^  (c  -^  d)     By  the  definition  of  a  fraction. 


6   •  (/ 

=  a  ^  b  -^  c  X  d 


Removing  parentheses  preceded  by  the 
sign  of  divison. 

=  a-^c^bxd  By  the  Commutative  Law  for  Division. 

=  (a  -f  c)  -f  (6  -^  fZ)     By  the  Associative  Law  for  Division. 


-Wd 


By  the  Notation  for  a  fraction. 


FRACTIONS  279 

Thut  is,  to  divide  one  fraction  by  a  second  fraction,  divide 
the  numerator  of  the  first  by  the  numerator  of  the  second  for  a  new 
numerator,  and  the  denominator  of  the  first  by  the  denominator  of  the 
second  for  a  new  denominatm\     (Compare  with  §  36.) 

Ex.  1. 
Ex.  2. 


35  .   7      35  -^  7      5 

48  •  8      48  -f  8  ~  6 

10a268  .  2a6_10a26s^2a6_5a62 
2\xhi    '    3a;  -  21a;2^  ^  3a;  "  Ixy' 

Check. 

a  =  6  =  2, 
x  =  y=:'^. 

This  process  is  useful  only  in  those  cases  in  which  the  numerator  and 
denominator  of  the  divisor  are  factors  of  the  numerator  and  denominator  of 
the  tlividend  as  shown  above. 

43.  The  reciprocal  of  a  fraction  is  unity  divided  by  the  fraction. 

That  is,  the  reciprocal  of  y  is  1  -r-  -r- 
b  b 

This  result  may  be  reduced  to  the  form  -• 

a 

The  fraction  bja  is  commonly  referred  to  as  being  the  reciprocal 
of  the  fraction  ajb. 

It  should  be  observed  that  the  reduced  value,  bja,  of  the  recipro- 
cal, 1  -7-  alb,  of  the  fraction,  a/b,  may  be  obtained  by  interchanging 
the  terms  a  and  b  of  the  given  fraction  a/b. 

The  product  obtained  by  multiplying  the  reciprocal  of  a  fraction 
by  the  fraction  is  unity. 

That  is,  ('lH-|)x|  =  l. 

44.  It  will  appear  upon  examination  of  the  expression 

c  d 

a  -^  -  =  a  X  - 

a  c 

that  the  operation  of  dividing  «  by  -3  has  the  same  effect  as  that  of 

multiplying  a  by  the  reciprocal  of  the  divisor,  that  is  by  -  (see  §  41). 

Hence,  for  the  operation  of  division  by  a  fraction  may  be  substituted 
that  of  multiplication  by  the  reciprocal  of  the  fraction. 


280  FIRST   COURSE   IN   ALGEBRA 

I>ivisiou  of  One  Fraction  by  Another 

45.  Let  it  be  required  to  divide  a/b  hy  c/d;  a,  b,  c,  and  d  repre- 
senting positive  integers. 

Applying  Principles  I.  and  II.,  §  34,  and  employing  a  course  of 
reasoning  similar  to  that  used  above,  we  obtain  the  result  as  follows : 

We  may  either  separate  the  fraction  a/b  into  c  equal  parts,  and 
then  take  d  of  these  parts  as  summands,  obtaining  ad/bc^  or  we 
may  substitute  for  the  operation  of  division  that  of  multiplication, 
using  the  reciprocal  of  the  divisor  as  a  multiplier,  and  immediately 

.,  a      c       a      d      ad 

b       a       b       c      be 

Hence  it  follows  that  the  quotient  obtained  by  dividing  one  fraction 
by  another  is  equal  to  the  product  of  the  dividend  and  the  reciprocal 
of  the  divisor. 

E     3      ?^^i£!^  =  ^^      ^h.  Check.     a  =  h  =  2, 

^'    '     4cd^  '  bbij  -4cd'^^2ch:  c=d  =  3,x  =  y  =  l. 

_(3a^6)(5%)  ^0  =  ^0. 

_  15  a^b^y 

2a2-f-7aft  +  6  62  a-h  a +  26 


Ex.  4.   Simplify 


ab-b^  ^  4a=2-962  • 


Factoring  and  substituting   x zr-,  for  -, j ,  we  niav  write 

^       a  +  2o  b 


9a^  +  7a6  +  6/>2       a-b        .rt+26_ 
ab-b^  ^4a2-962^       l       - 

(2q  +  .3  6)(a  +  2Z>)      a-b b 

b{a-b)  ^  (2a4-36)(2«-36)  ^  a  +  26 

_         1  Check.     « =  4,  b  =  2. 

=  2a-36 '  ^  =  1 

Since  the  factors  are  removed  from  the  numerator  by  division,  not  by 

subtraction,  it  follows  that  when  we  strike  out  the  last  factor,  whichever 

that  may  be,  the  quotient  1  remains. 

46.  Mixed  numbers  appearing  in  either  dividend  or  divisor 
should  be  reduced  to  fractional  form  before  the  division  is  carried 
out. 


FRACTIONS  281 

Ex.  5.   Divide  -^  +  1  +  -^  by  -^^^^ 
a  +  6  a  —  b    "^  a^  —  b^ 

Expressing  the  dividend  as  an  improper  fraction  and  multiplying  by 
the  reciprocal  of  the  divisor,  we  have 

/    a  b    \  _^      a^      _  gg  -  aft  +  gg  -  52  +  a6  +  b^      oP-  -  b"^ 

'  1  +  6  "*"     "^  ^^^>y  "^  a2  -  62  -  a2  -  6^  x  — ^^ 

_  2a2  Check,    a  =  3,  6  =  2. 

~  a^  2  =  2. 


Mental  Exercise  XV.     10 
Express  each  of  the  following  quotients  in  simplest  form 


1.  -r  -i-  a. 
0 

12. 

-*''■ 

_    5m 

2. '-m. 

n 

13. 

K-- 

«    12a;      ^ 

3.  —20;. 

3^ 

14. 

-H 

4.  l^«.(-3o). 

15. 

1 

a.?.. 

16. 

5 

6j^(-.). 

17. 

..^^. 

18. 

3-'- 

8.  |j^3a. 

19. 

a  .   2 
b  '   3' 

9.1^26. 

20. 

a;  .   4 
2/   •   5' 

-¥-• 

21. 

7  .  m 

8  '  ^' 

ll.V*^(-5). 

22. 

9  .  d  ^ 

10  *  c  ' 

23. 

1 

2   * 

X 

y' 

24. 

1 
3  • 

a 
'V 

25. 

m 
n 

1 
'   4' 

26. 

w 

-{-\ 

27. 

a 

b  ' 

d 

c 

28. 

X 

y 

.1. 

X 

29. 

m 
■7 —  • 

n 

30. 

b^~ 

.  b^ 
a 

31. 

ab  ^ 
c 

c 

32. 

X 

yz 

X 

33. 

abc 

ab 

xy 

xyz 

282  FIRST  COURSE   IN  ALGEBRA 

Exercise  XV.     11 

Express  the  following  quotients  as  single  fractions  in  lowest 
terms,  checking  all  results  numerically  : 

'jji^db_  _    6  —  6  .       3 

*  36   *   14a' 


2. 

2a^ 
56    • 

8  a* 
15  6* 

3. 

9a6«  . 

12a«6 
3b  xy^ 

4. 

dx 

Uz 

lOy  ' 

12  w 

5. 

5ci/ 

4:bZ 

Saw 

6. 

12  ad 

32  be 

23  Oc 

'  21  ad 

7. 

3a%  . 

4ab^ 
3cH 

8. 

1 
\-x 

1 

•     l+iC 

9. 

X—  1 
X 

.     y 

'   x^-  1 

10. 

«+ 1 

5 

*     10a 

11. 

a-\-  1 
2 

3 
•  a+1 

12. 

x  —  2 
3 

5 

'  x  +  2 

13. 

m  -\-  n 

8 

6 

'   m  -\-  n 

1  4 

a  +  2 

1 

a  +  3. 


xo» 

10      '6+1 

16. 

cH-4  .       4 
5       '  c  +  b 

17. 

a-\-b  ,  SB  — y 
X  -\-  y  '  a  —  b 

18. 

10a'           5a 

(a  +  by  '  a  +  b 

19. 

9  a:'      .      3x 
a;' -9  *  a; -3 

OA 

18a;       .       6aj 

16a;'- 1   •  4a;- l' 

21. 

2          .           4 

2x  +  3ij  '  4a;'-92/' 

22. 

c'  -  9       .        c  -  3 
c'  —  c  -  2  *  c'  +  c  —  6 

23. 

w'  —  w  —  2   .  m^  —  2m 

TV?  —  m  —  ^   '   2  m  -{•  m^ 

24. 

a-3          .  a^  -  9 
a'  — 2a +  4  *  a' +  8 

25. 

(a;  +  ?/)'  .   (a;'  -  y^ 
x+3y    '  x^  ■\-21f 

26. 

33       .            11 

a«  +  6«  •  d'-ab-\-  b^ 

27. 

x^  —  f  ^  x^  +  xy  -{-  y^ 
14       •              7 

28. 

a«  +  1     .  a'  -  a  +  1 

^2__4    a          9J-2V 

FRACTIONS  283 


\c      ah)       \a      b       cj 

ob.    — -— X  — s r   X 


37. 


Qax  —  2()x        a^  — 9       3a -8   '   'da'  —  a  —  ^O 
{a  +  by  -  if  a  ^    ab  -  by  -  b^ 

a^-aij  +  ab       (a  +  yY  -  b^  '   (a-  bf  -  f 


m^-Vm-^      7??H8m+15  .  ^m'^  +  Am  +  S      m  +  S\ 
m^  —  m  —  SO     m^  —  4m  +  4:  '  \m^  —  4:m  —  V2     m—  2/ 

Complex  Fractions 

47.  Since  in  algebraic  expressions  we  cannot  restrict  ourselves  to 
using  positive  integers  only,  we  shall  find  it  necessary  to  admit  to 
our  calculations  iractions  whose  numerators  and  denominators,  either 
or  both,  contain  terms  which  in  themselves  may  be  positive  or 
negative,  integral  or  fractional. 

By  the  Principle  of  No  Exception  we  shall  so  extend  our  ideas 
concerning  fractions  that  whatever  may  be  the  nature  of  the  numbers 
entering  into  the  terms  of  any  particular  fraction,  the  operands  must 
in  all  cases  be  governed  by  the  laws  demonstrated  for  fractions  whose 
terms  consist  of  positive  whole  numbers  only. 

48.  A  complex  fraction  is  a  fraction  containing  one  or  more 
fractions  among  the  terms  of  its  numerator  and  denominator. 

E.  g.   The  following  expressions  are  complex  fractions : 
a-\-b  1  _  1 

a  c  ah 

T'  rf~'       c+d' 

c  ab 

49.  When  simplifying  a  complex  fraction  it  is  often  convenient 
to  obtain  the  reduced  value  of  the  numerator  alone  and  of  the  denom- 
inator alone  before  attempting  to  perform  the  indicated  division. 


284  FIRST  COURSE  IN  ALGEBRA 

Ex.  1.  Simplify 


1  _  1 
a b_ 

L_L 

32      62 


We  have. 


60.  It  is  sometimes  desirable,  when  simplifying  complex  fractions, 
to  multiply  both  the  numerator  and  denominator  by  the  L.  C.  M.  of 
the  denominators  of  the  fractional  terms. 

E.  g.  In  the  example  above  we  might  have  multiplied  the  complex  num- 
erator and  also  the  complex  denominator  by  the  L.  C.  M.  of  their  denom- 
inators,  a%^,  and  have  written 


xa%^ 


ab       _      ab  _  (^  —  a)«^  _     «& 


62  _  rt2  -   b-i-  «2  -  62  _  a2      -  a  +  6 

— ^TT-        — 212-  X  *^ 
a^b^  a^b^ 

It  will  be  observed  that  in  this  particular  instance  we  have  not 
used  the  reciprocal  of  the  divisor  as  a  multiplier,  but  have  adopted 
another  method. 


Ex.2. 


a  +  b      a  —  b         {a  +  b      a-b\ 

a-b      a  +  b  _   [a-b      «  +  ij^"  +  *>("  " 

-6) 

a  +  b      a  —  b   ~    /a  -\-  b      a  —  b\^         ^., 
a-b^a  +  b        (a-6  +  a4-6r  +  '^^^- 
(a  +  6)2  -  (a  -  6)2 

-  (a  +  by  +  (a  -  6)2 
_    2a6 

-  a2  +  62 

-6) 

Check, 
a  =  3,  6  =  2. 
12      12 
13~13* 

Exercise  XV.     12 

Simplify  the  following  complex  fractions,  checking  all   results 
numerically : 


FRACTIONS  285 

1  +  1  --  +  '- 

4.  —  •  7.  ^ 


^+^  1-i  ^-l 

z  y  a      z 

2.  _l^.  5.  ^.  8.  — ^±1- 

C  TTt  X  —\ 

1,1  1_1  ^_2/_^ 

3    .^ 2^.  g   _f^ y_.  ^    y      z      Ic 


^      y  ^      f  y      z      X 

14                             a  +  ^.«  —  ^ 
<^_2 -—  -H —- 

-^                    a  +  3  IK     c  — »      c  +G? 

10.  — 15. 


21  a  +  b  ,  a  —  b 

a  ■\-  6  c  +d      c  —a 

1- _— —  c— 3 c— 4 


11.  .       ^  +  ^»  16.  ^X ^ 


b       c  c— 4  c— 1 

a  +  t*  H —  c         a  ,     , 

12.  ^.  17.         ^  •^+^- 


b  y      c 


13. 

1      +      ^ 

1  — ic       1  +  a 

1              1 

l—x      1+a 
^        2b-2c 

14. 

a  +  b  —  c 

1+     \' 

1_]_         •   c'-y'' 
y     c 

1  +  2^       1-2^ 


1-2^       1  +  2^- 

'    1  —  2^       1  +  2  /; 

1  +  2^"^  1  —  2A; 

i  +  l  +  l 

a;?/      a;^      y^ 
^^-    a;^  -  (y  +  ^)^  ' 
a;y 


286  FIRST  COURSE  IN  ALGEBRA 

ah        ah  oo^  2w  —  I 


h      a    ,    b  a                     '  w  +  I          2,«^1 

a^6          1  ,1                                                  ^22 

0      a        a  0 

X       y  y^\                        1 

_  _l_  Z.  23 I . 


a^      ary 


t       x  +  y-r. 


24.   — ^ +  2 

y  aj  2/      a^ 


Continued  Fractions 


51.   Rational    functions    of   the    form 


are  commonly  called  continued  frac-       i  j^ ^ 

tions.  7  ,       e 

52.   Certain   simple   forms   of  con-  f  A-i. 

tinued  fractions  may  be  simplified  by  h 

beginning  at  the  last  mixed  expression, /+  |,  and  after  reducing 

this,  by  proceeding  upward,  reducing  successively  the  next  higher 
complex  and  mixed  fractional  expressions,  until  finally  all  have  been 
used. 

2 


Ex.  1.    Simplify 


1- 


-? 


/  2 1, 

/ — -^  ■     ^-^  " 


',^         In  the  accompanying  figure  we  have  indicated  the 

^ /^  _^^  J*    ^^-^  successive   steps   of  the  process,   1,   2,  3,  and   4,  by 

I  ',         1/5   4.  Ay  ly  enclosing  the  different  mixed  and  complex  fractional 

\\^     ^^^  -i  -^S/  expressions  in  different  "  spaces  "  bounded  by  curved 

^^•^s-^-''^'^  lines. 

The  fraction  may  be  simplified  as  follows: 


o 

**" 

41 
6 

'j8. 
41 

—  __  , 

\ 
\ 
\ 

/ 

^ 

2^ 

41 

-'3^- 

= 

41 

/ 
/ 

\ 

2 

23 
41 

\ 
\ 

J 
/ 
/ 

= 

_82. 

.        41         1 

23 

287 


By  reducing  the  improper  fraction  ||  we   obtain   the  mixed  number 
3J|,  which  is  the  value  of  the  given  continued  fraction. 

Ex.  2.   Simplify  the  following  continued  fraction  ^ — . 

Simplify  the  lowest  mixed  expression  and  di-  ^ ^ 

vide  h  1  >y  the  result.  Subtract  the  result  of  this  oper-  a 

ation  from  c.     Use  the  reciprocal  of  the  remainder 

as  a  multiplier  of  a,  to  obtain  the  reduced  value  -r ^  of  the  given 

continued  fraction. 

Check,  a  =  4,  ft  =  3,  c  =  2. 
5  =  5. 
Exercise  XV.     13 

Simplify  each  of  the  following  : 

1                    .           «  „  a  —  2 

4. 7. 


X. 

1 

a 

+ 

— 

1 

a 

+ 

a 

a 

2. 

h 

X 

+ 

— 

c 

y 

+ 

z 

a 

a. 

h 

b 

— 

— 

a 

a 

—~ 

h 

h a  — 2 


5.  1 


e  a  — I 

1  .  1 


1  m'—l 

1 m 


1-1  m+       ' 


X  m  —  1 


-! 


288  FIRST  COURSE  IN  ALGEBRA 

53.  Fractions  whose  terms  contain  binomial  sums  and  differences 
may  be  simplified  as  follows  : 

h  -{-  c  a  -\-  c  a  +  b 


Ex.  1.   Simplify 


(a  _  b){a  -  c)  ^  {b  ;^(6  -a)^  {c-  a)(c  -  6) 


Among  the  six  binomial  factors  oi  the  denominators  there  are  but  three 
which  are  essentially  different,  such  pairs  aa  a  —  b  and  b  —  a  differing  only 
in  sign. 

We  may  accordingly  choose  for  the  form  of  the  L.  C.  D.  the  expression 
(a  —  b)(b  —  c^)(c  —  a).     Hence  we  may  write : 

6  +  c  a  +  c  a  -\-b 


(a-6)(a-()   '    (/,  _c)(6-a)   '   (c  -  a)(c  -  b) 

=      -  (^  +  0       .       -  («  +  0        ,       -  (a  +  b) 

-  (a  -  6)(c  -  a)  "^  (6  -  c)(a  -b)'^  {c-  a)(b  -  c) 

(The  —  signs  are  written  in  the  numerators  to  compensate  for  the  change 
of  sign  resulting  from  the  alteration  of  the  signs  of  the  factors  in  the 
corresponding  denominators.) 

_-(h  +  c)(b  -c)  -(C  +  aXc  -a)  -(a  +  b)(a  -  b) 

ia-bXb-jXp-^) 
_-b^-\-c^-c^  +  a^-  a^  +T^ 

-  (a  -  b)(b  -  c){c  -  a) 

(It  will  be  seen  that  the  mutually  destructive  terms  in  the  numerator 
disappear  through  addition  and  subtraction,  for  example,  a^  —  a^  =  (). 

Hence,  when  striking  out  the  last  pair,  whichever  that  may  be,  we  must 
write  0  as  a  resulting  numerator.) 


-  (a  -  6)(6  -  c)(c  -  a) 

=  0,  providing  a  :^  b  ^  c.         Check,     a  =  4,  ft  =  3,  c  =  2. 

0  =  0. 

This  is  because  the  value  of  a  fraction  is  0  if  the  numerator  be  0  and  the 
denominator  different  from  0. 

54.   Complex  Fractions  Involving  I>ecimal  Fractions. 

Although  no  new  principles  are  introduced  by  the  appearance  of 
decimal  fractions,  much  labor  may  be  saved  by  means  of  certain 
special  devices. 

E.  g.  By  expressing  all  of  the  decimal  fractions  and  whole  numbers  in 
a  complex  fraction  as  decimal  fractions  of  the  same  order,  we  may  disregard 


FRACTIONS  289 

the  decimiil  points  altogether,  since  this  amounts  to  multiplying  both 
numerator  and  denominator  of  the  complex  fraction  by  the  number  repre- 
senting the  order  of  the  decimal  fractions. 

E     2  3.06  g  +  -0204  _  30600  a  +  204 

.255  ~  2550 

_  300a +  2    .      ,     ,     .    „ 
•     Check,     a  =  2.  =  _^-_ ,  m  algebraic  form, 

24.08  =  24.08.  =  12  a  +  .08,  in  decimal  notation. 

Exercise  XV.     14     Miscellaneous 

Simplify  the  following  fractional  expressions,  checking  all  results 
numerically : 

1.  1-a  +  a  -j-T—'  6.  -— ^ 


1  +  a  3a'  7 

2.  x^  +  xi/  +  y'  +  -^ 7.  -2 T-^  X       ,   , 

x  —  y  m^  —  ^x'        imr  -\-  mn 

a^b  -  h^             Sa  ^-4       d' -  1       d-2 

a             2ab-2b'''  ^'  d' -  l^     2d      ^2  +  d 

^   m^-a'  ^^m^  +  a\  ^    /I        IV    a^>    \ 

az            a  —  m  '   \d^      b^ )\a  +  b) 

x  +  y       xy  —  y^  \  xj\x  —  yj 

\x       a  J\x^      a^J 

4.  o  9 

12.    7 —. ^  + 


{a  —  l)(a  —  H)       (a  -  2)(3  —  a)       (2  —  «)(!  -  a) 
2^a  +  b  —  c2^a  +  c  —  b2      b  -\-  c  —  a 
a  ab  b  ac  c  be 

14    ^''-2^+1       F-4^+  4      Jc'  -6^+9 

,^     3a2  +  «-24         6a'          a- 3  3«'  +  9a 

15.   — — ■ X  —^ X X 


6  aa;  —  20  ic        a'  —  9       3  a  —  8       3  a'"^  —  a  —  30 
V  a;+«y\  X  —  a  J 


19 


290 


FIRST  COURSE  IN  ALGEBRA 
4 


18.    M^       I    1  Mr      ^        I    1  ^^     V 

'    \2  +  a  a     J  '  \a  +  2  a     J 

m  +  n         \m  —  n      m  -\-  it) 
21.  ii  I      "^    V;^~"'%(      "'^^  ^ 

\         m  —  ^J       4  —  w*      w^  +  w  —  6      7W  +  2 


23. 
24. 
25. 
26. 


a^bc 


h^ca 


+ 


c^a6 


(a  —  b){a  —  c)       (b  —  c){b  —  a)       (c  —  a){c  —  b) 

r+\  s+l  t+l 

^r-.s){r-t)^  {s-t)(s-r)'^  {t-r)it-s)' 


(c"  -  d')((^  -  e^)   '   (cP  -  e^)(d' -  c^)   '   (d' -  c'Xe'' -  (P) 


d  +  b  —  c 


c  +  d-b 


b  +  c  —  d 


(d-b)(d-c)   '   (c-d)(c-b)   '   (b-€){b-d) 


Indeterminate  Forms 

Sections  55-72  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 

55.  It  frequently  happens  that  when  particular  values  are 
assigned  to  the  letters  appearing  in  the  terms  of  a  fraction,  either 
the  denominator  or  the  numerator  or  both  become  zero. 


E.g. 


a:+  1 


X  +2 


3C^  —  X^ 


for  a:  =  1  becomes 

for  a:  =  —  1  becomes 

for  ar  r=  2  becomes 

for  X  =  —  2  becomes 

for  X  =  0  becomes 

for  X  =z  I  becomes 


FRACTIONS  291 

For  all  other  finite  values  of  the  letters  the  expressions  above  assume 
perfectly  definite  values. 

56.  According  to  definitions  previously  given,  expressions  in 
which  zero  appears  as  a  divisor  have  been  excluded  from  calcula- 
tions as  being  meaningless  as  numbers. 

In  order  that  such  "  number  forms  "  may  without  exception  be 
admitted  to  our  calculations,  we  proceed  to  extend  our  idea  of  the 
value  of  an  expression. 

Since  Quotient  X  Divisor  =  Dividend, 

we  may  regard  8  as  meaning        Quotient  X  0  =  0. 

Hence,  since  the  product  of  any  number  and  zero  is  zero,  we  may 
interpret  §  as  representing  any  number. 

57.  If  a  variable  be  supposed  to  change  in  value  in  such  a  way 
as  to  become  and  remain  as  nearly  equal  as  we  please  to  some  definite 
fixed  value  or  constant,  the  variable  is  said  to  approach  the  constant 
as  its  limit. 

The  symbol  ==  is  read  "approaches  as  a  limit." 

E.  g.  The  expression  ^^^^  (x  +  6)  =  a  +  &  is  read,  "  the  limit  of  {x  +  h) 
as  X  approaches  a  is  equal  to  a  +  &." 

a:2  _  25  0 

The  fraction —  assumes  the  form  -  when  x  —  6. 

X  —  b  0 

We  may  write  ^-^  =  /_%  =  i- +  "^^^  ' 

If  we  suppose  that  x  approaches  5  as  a  limit,  then  so  long  as  x  has  a  value 

diflFerent  from  5, will  have  the  value  unity  (since  the  numerator  and 

a:  —  5 

denominator  are  equal),  while  x  +  5  will  differ  from  10  by  exactly  that 
value  by  which  x  differs  from  5. 

Accordingly  we  can  make  the  value  of  {x  +  ^)\~Z^]  ^^  nearly  equal 
to  10  as  we  please  by  giving  to  a;  a  value  sufficiently  near  to  5. 

Accordingly,  ^^,  (^if^)  =  /f^  [(-  +  ^^i^^^^]  =  ^^- 

58.  We  shall  define  the  value  of  an  expression  for  any  par- 
ticular value  of  its  variable  to  be  the  limit  (if  there  be  one)  ap- 
proached by  the  expression  as  its  variable  approaches  the  particular 
value  as  a  limit. 


292  FIRST  COURSE   IN  ALGEBRA 

Although  this  general  definition  may  be  used  in  all  cases,  we 
shall  employ  it  only  when,  by  using  the  definition  given  in  Chap.  I., 
§  1 9,  we  fail  to  obtain  a  definite  number. 

59.  To  find  the  value  of  a  fraction  which  assumes  the 
indeterminate  form  0/0 /w  some  ^particular  value  of  the  variable 
appearing  in  it ^  find  the  limit  approached  by  the  fraction  when  the 
variable  approaches  the  given  particular  value  as  its  limit. 

60.  It  should  be  observed  that  a  rational  fraction  assumes  the 
form  0/0  because  some  factor  common  to  both  numerator  and  de- 
nominator becomes  zero  for  some  particular  value  of  the  letter 
appearing  in  it. 

61.  Since  the  symbol  0/0  does  not  represent  the  same  value  all 
of  the  time,  but  assumes  different  values  according  to  circumstances, 
we  interpret  0/0  as  representing  an  iiideteriiiinate  value. 

Ex.  1.    Find  the  limiting  value  of  the  fraction  {x^  —  6  a:  +  9)  /(a:  —  3) 

w~              .,    a:2_6a;  +  9_.            /x-3\ 
We  may  wnte —  =  (x  —  3)  I  ^  )  • 

I^OT  all  values  of  x  different  from  3,  ar  —  3  ^t  0. 

a:  —  3 

Hence  the  value  obtained  by  multiplying  a:  —  3  by -,  wliich  is  unity 

X  —  o 

when  X  is  different  from  3,  is  the  value  of  the  factor  a:  —  3. 

As  X  approaches  3  as  a  limit  the  expression  a:  —  3  approaches  zero  as  a 
limit. 

Accordingly  the  value  of  the  given  fraction  is  taken  as  zero  when  a:  d=  3. 

Another  method  for  finding  the  limiting  value  of  an  indeterminate 
fraction  is  shown  in  the  following  example  : 

Ex.  2.   Find  the  value  of  {x^  -  49)/(x  +  7)  when  x  =  -  7. 
We  may  indicate  that  x  differs  from  —  7  by  writing  x  =  —  7  +  hj  letting 
h  represent  a  value  which  may  be  made  as  small  as  we  please. 

jp2 49 

Accordingly,  substituting  h  —  7  for  x,  '  becomes 

X  -p  I 

(h  -  ly  -  49  _  /62- 14/1 +  49 -49 
{h~l)  +  l    "=  h-1  +  1 

=         h        ' 
for  all  values  of  /i  ^t  0,  =h  —  14. 


FRACTIONS  293 

Hence  (,/;-  —  4Q)/{x  +  7)  differs  from  —  14  by  li,  which  may  become  and 
remain  as  small  as  we  please. 

Accordingly  the  given  fraction  approaches  —  14  as  a  limit  as  a?  approaches 
-7. 

62.  Consider  the  fraction  ajx  in  which  the  numerator  a  is  re- 
garded as  having  some  fixed  or  constant  vahie,  different  from  zero, 
while  the  value  of  the  denominator  is  subject  to  change. 

As  the  denominator  is  given  successively  smaller  and  smaller 
values  (1/10,  1/100,  1/1000,  etc.),  the  numerator  retaining  some 
constant  value,  it  may  be  seen  that  as  the  value  of  the  denominator 
decreases,  the  value  of  the  fraction  increases. 

E.  g.  -^  =  10  a,     -^  =  100  a,     -^  =  1000  a,  etc. 

By  giving  to  the  denominator  a  value  small  enough,  the  value  of  the 
fraction  a/ x  can  be  made  greater  than  any  assignable  number. 

63.  The  symbol  x> ,  read  "  infinity  ",  is  commonly  used  to  denote 
all  numbers  or  values  which  are  greater  than  any  assignable  arith- 
metic number  or  value. 

The  expression  x  =  y:),  read  "a;  increases  in  value  without  limit " 
or  "  X  is  infinite  ",  is  to  be  understood  as  meaning  that  x  has  no 
definite  fixed  value,  but  that  it  may  assume  dij^ereMt  values  which 
are  very  great  and  are  beyond  the  range  of  computation  or  imagi- 
nation. 

64.  The  symbol  for  an  infinite  number,  oo,  should  never  be 
treated  as  representing  a  definite  value,  and  it  is  not  subject  to  the 
Laws  of  Algebra. 

Corresponding  to  the  ideas  of  positive  and  negative  numbers,  we 
have  "*'ao ,  read  "  positive  infinity,"  and  "oo ,  read  "  negative  infinity." 

65.  It  is  impossible  to  separate  unity,  or  in  fact  any  number,  into 
such  small  parts  that  one  of  these  parts  shall  have  no  value  at 
all,  that  is,  be  zero.  Hence  it  may  be  seen  that,  although  the 
successive  denominators  (1/10,  1/100,  1/1000,  etc.)  of  the  complex 
fractions  in  §  62  become  smaller  and  smaller  in  value,  we  can  never, 
by  diminishing  the  denominators  in  this  way,  obtain  zero  as  a 
denominator. 

66.  A  variable  whose  value  may  become  indefinitely  small  with- 
out ever  becoming  zero  is  called  an  infinitesimal. 


204  FIRST  COURSE  IN  ALGEBRA 

67.  The  symbol  o  (horizontal  zero),  —  read  "  diminishes  inde- 
finitely in  value  without  ever  becoming  zero,"  or  "  is  indefinitely 
or  infinitely  small,"  or  "is  nearly  zero,"  —  has  been  used  by  certain 
writers  to  denote  an  infinitesimal  variable. 

E.  g.      X  ■=.  o  means  "  is  nearly  zero,"  that  is,  has  a  very  small  value. 
X  =  0  means  "  is  exactly  zero,"  that  is,  has  no  value. 

68.  Such  numbers  as  are  neither  infinite  nor  infinitesimal  in 
value  are  called  finite  numbers. 

69.  Interpretation  of  g.  It  may  be  seen  from  the  preceding 
paragraphs  that  if  the  numerator  of  a  fraction  remains  constant  in 
value,  the  value  of  the  fraction  as  a  whole  increases  when  the  value 
of  the  denominator  decreases ;  that  is,  for  a  given  numerator,  the 
smaller  the  denominator  the  greater  the  value  of  the  fraction. 

Hence,  as  the  denominator  becomes  infinitesimal  the  fraction 

becomes  infinite.     Hence  we  may  write  ^  =  qo  . 

Accordingly,  although  strictly  speaking  -  has  no  meaning,  we 

shall  define  it  to  have  the  same  meaning  as  -^ ;  that  is,  we  shall 
interpret  -  =  oo  • 

70.  Interpretation  of  ^ .    It  may  be  seen  that  if  the  numerator 

of  a  fraction  remains  fixed  in  value,  the  value  of  the  fraction  as  a 
whole  becomes  smaller  indefinitely,  as  the  value  of  the  denominator 
increases  indefinitely. 

Accordingly  we  may  write  ^  =  ^  * 

It  should  be  observed  that  -^  is  defined  to  mean  nearly  zero^  that 
is  o,  not  exactly  zero^  which  is  0. 

71.  By  the  Principle  of  No  Exception  we  shall  define  -  to  mean 

^,  (which  may  have  any  finite  value),  -,  to  mean  ^,  that  is  qo, 

(which  represents  any  infinite  value),  and  ^  to  mean  o  (an  infini- 
tesimal value). 


FRACTIONS  295 

The  expressions  - ,  - ,  ^  are  commonly  called  indeterminate 

forms. 

72.  The  symbol  ]s ,  written  at  the  right  of  a  fraction  or  other 
expression  containing  a  single  variable,  is  to  be  understood  to 
denote  that  the  value  represented  by  the  subscript,  s,  is  to  be 
substituted  for  the  variable. 


■n,  rru  .       a;  +  n  3+14 

E.  g.     The  expression  ^-^ J  ^  means  ^-^  =  - 


Exercise  XV.     15 

Find  the  limiting  values  of  the  following  expressions  for  the 
values  specified  : 


*  a;  —   1  Ji  *       ic  —  2      Ja 

"-25"|  x^-4:x''  +  x+  G") 


1 

a--25 
'  a  +  5 


3.^  + 


4-  n  a;'-3a;'^  +  3^-  l"[ 

+  lj-i'  '  i«-l  Ji* 

6^-1"!  a;«-6a;'^+  12a; -8"| 

6  -  iji*  *  a;'-4a;  +  4        Ja* 

Factors  of  Fractional  Expressions 

73.  Expressions  which  are  fractional  with  respect  to  specified 
variables  may  be  factored  by  the  methods  which  are  employed  for 
factoring  integral  expressions. 

a      a       I         (        1\       1/        1\  Check. 

Ex.2.     _-_..(^-  +  -j(^---j. 


Check. 
Let  ft  =r  3,  &  =  2,  c  =  1. 

17  _  11 
115  —  16- 


296  FIRST  COURSE  IN  ALGEBRA 

Exercise  XV.     16 
Factor  the  followiDg  expressions,  checking  all  results  numerically : 


a^      ab      b^ 

7. 

6«^       13« 
b'    ^    b     ^^' 

-«^^¥n- 

8. 

a^              a 

3..y-i. 

9. 

a      or 

4.  -«  —  2+  -^• 

10. 

a^  +  a  +  2  +  i  +  i 
a      a^ 

.$,%.. 

11. 

1       2/« 
ic«       8* 

y      y 

12. 

(^O-'CT 

Mental  Exercise  XV.  17     Review 

Obtain  the  following  products  : 

1.  (x  -  1)(1  -  x),  3.  (4  +  d){d  -  4). 

2.  (2  -  h)(b  -  2).  4.  (c  +  5)(-  c  -  5). 
Simplify  each  of  the  following  : 

5.  1  -r-«/6.  7.  1  -=-  1/^y. 

6.  rt  -h  1  /6.  8.  xy  -^  a;/2/. 
Distinguish  between 

9.  -  +  -  and  — ; 10. y  and =- 

^      y         X  +  y  a      0         a  —  0 

11.  __and-+l. 
Simplify  each  of  the  following  : 

6  +  c      b  —  c  y  —  z      y  +  z 

Supply  the  terms  which  make  the  following  expressions  trinomial 
squares  : 

14.  a^+8a+(    )•    17.  ^«+6^4-(    )•     20.  9A*+6^«  +  (    ). 

15.  P—lOb+l     ).     18.  m^  +  2m^-\-(     ).     21.  16£c«-24a;'+(     )• 
IQ.  c*  + 2(^+1    ).    19.  w*+4?i*  +  (    ).     22.  25y^'-S0f+l    ). 


FRACTIONS  297 

Are  the  following  expressions  conditionally  or  identically  equal  ? 

23.  6  +  2  and  5-1-3.  25.  6  a;  +  2  and  5  ic  -h  3. 

24.  6x+  2x  and  5  -h  3.  26.  6 a;  -h  2 aj  and  5 a;  -f-  3 a. 

27.  To  which  of  the  equations,  5a;+  2y  =  1,  3a;  —  4^/=  8, 
4  a;  --  3  ?/  =  5,  are  the  following  equations  equivalent  ? 

10x+  4y=U;     lox+  Qy  =  21;     6x-8y  =  lQ', 
9  a;—  12  y  =  24;     Ux—12y  =  20. 

Are  the  following  expressions  identical  ? 

28.  -Ti  ^  bhy  -  /*,  and  b  -  • 

29.  ^,|+|andi(.:  +  j,). 

Show  that  the  following  identities  are  true  : 
1  1 


30. 

{b  ^  a){c  -  b)  —  {a  -  b){b  - 

■c) 

31. 

1        _        1 

{d  -  cf  -  (c  -  dy 

<?<> 

1            1 

(b-ay-(a-by 

33. 

(«-.)^_^    .y. 

298  FIRST  COURSE   IN  ALGEBRA 


CHAPTER  XVI 

FRACTIONAL  AND  LITERAL  EQUATIONS 

Equations   which   are  Algebraically   Rational  and 

Fractional  with  Reference  to  a  Single  Unknown 

Having  Numerical  Coefficients 

1.  An  equation  is  said  to  be  fractional  with  respect  to  any 
specified  letter  if  that  letter  appears  in  the  denominator  of  a  fraction 
in  either  member  of  the  equation. 

2x5 
E.  g.  The  equation H =  7  is  fractional  with  respect  to  a;, 

Q  ~  Ax  h 

while r  —  ;7—  = r  is  inteffral  with  respect  to  x  but  fractional  with 

a  +6      2a      a  —  6  ^  ^ 

respect  to  a  and  h. 

2.  The  only  equations  which  are  spoken  of  as  having  degree  are 
those  which  are  entirely  rational  and  integral  with  respect  to  the 
unknowns  appearing  in  them. 

Hence  the  tenn  degree  does  not  apply  to  equations  which  are 
irrational  or  fractional  with  reference  to  a  specified  unknown. 

3.  If  a  given  fractional  equation  cannot  be  solved  immediately 
by  inspection,  its  solution  may  be  made  to  depend  upon  the  solution 
of  an  integral  equation  derived  from  it  by  multiplying  both  members 
by  the  lowest  common  denominator  of  all  of  the  fractions  appearing 
in  it. 

This  process  of  deriving  an  integral  equation  from  a  fractional 
equation  is  spoken  of  as  clearing  the  fractional  equation  of 
fractions. 

4.  The  equivalence  of  the  given  fractional  and  the  derived  inte- 
gral equations  may  be  determined  by  the  following 

Principle:  If  both  members  of  an  equation  whose  terms  are 
rational  and  fractional  with  reference  to  a  single  unknown^  x,  be 
multiplied  by  suck  an  integral  function  of  x  as  is  necessary  to  clear 


FRACTIONAL  EQUATIONS  299 

the  members  of  fractions^  then  the  derived  integral  equation  will 
he  equivalent  to  the  given  fractional  equation. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  the  terms  of  a  given  equation  which  is  rational  and  fractional  with 
reference  to  a  specified  unknown,  a:,  be  all  transposed  to  the  first  member, 
and  then  added  algebraically. 

Let  the  resulting  fractional  first  member,  reduced  to  lowest  terms,  be 

N 
represented  hy  -jz,  in  which  N  and  D  represent  expressions  which  are 

rational  and  integral  with  reference  to  the  unknown  x. 

Then,  by  the  principles  of  equivalence,  the  given  fractional  equation 

N 
will  be  equivalent  to  the  derived  fractional  equation  —  =  0.     (1). 

N  .    . 
Since  y:  is  in  lowest  terms,  it  follows  that  N  and  D  have  no  factor  in 

common. 

If  for  any  value  of  x,  such  as  x  =  a,  both  N  and  D  should  become  zero, 

it  would  follow  from  the  Factor  Theorem  that  N  and  D  would  have  the 

N 
factor  a;  —  a  in  common,  and  accordingly  the  fraction  ^  would  not  be  in 

lowest  terms. 

Accordingly,  when  for  any  particular  value  of  x  the  numerator  N  of  the 

N 
fraction  -tt  ,  which  is  in  lowest  terms,  becomes  zero,  the  denominator  D 

must  be  different  from  zero. 

.       N  . 
The  necessary  and  sufficient  condition  that  the  fraction  -y-  in  lowest 

terms  shall  become  zero  is  that  the  numerator  N  shall  become  zero,  the 

denominator  D  remaining  finite  and  diff'erent  from  zero. 

N 
Hence  any  solution  of  the  fractional  equation  -tT  =  0  (2)  must  be  a 

solution  also  of  the  integral  equation  iV=  0  (3),  obtained  by  multiplying 
both  members  of  the  fractional  equation  (2)  by  the  multiplier  D,  which  is 
necessary  to  clear  equation  (2)  of  fractions. 

Hence  no  solutions  of  the  fractional  equation  (2)  are  lost  by  multiplying 
both  members  by  the  multiplier  D. 

Since  any  value  of  x  which  reduces  iV  to  zero  cannot  reduce  D  to  zero 

also  (because  N f  D  is  in  lowest  terms),  it  follows  that  any  solution  of  the 

integral   equation  N  =  0  (3)  must  be  a  solution  also  of  the  fractional 

N 
equation  77  =  0  (2). 


300  FIRST  COURSE   IN   ALGEBRA 

Hence  no  solutions  are  gained  by  multiph'ing  the  fractional  equation  (2) 
by  the  multiplier  D. 

Accordingly,  the  given  fractional  equation  and  the  derived,  integral  equa- 
tion N  =0  (3)  are  etjuivalent. 

5.  Although  an  integral  equation  which  is  equivalent  to  a  given 
rational  fractional  equation  may  be  derived  by  transposing  all  of  the 
terms  of  the  fractional  equation  to  the  first  member,  uniting  these 
terms  into  a  single  fraction,  reducing  this  fraction  to  lowest  terms, 
and  then  clearing  the  ecpation  of  fractions,  it  is  not  always  conven- 
ient to  carry  out  the  steps  of  the  process  in  this  order. 

6.  We  shall  consider  in  this  chapter  rational  fractional  equations 
containing  a  single  unknown,  the  solutions  of  which  may  be  made 
to  depend  upon  the  solutions  of  linear  equations. 

Other  fractional  equations  will  be  discussed  in  a  later  chapter. 

11  9 

Ex.  1.    Solve  the  fractiqnal  equation =  - — '- — —  •  (1) 

X  -j-  1         Ji  X  —  1 1 

We  may  derive  an  iutegral  equation  by  multiplying  both  members  of 
(1)  by  the  protluct  (x  +  l)(2a;  —  11)  of  the  denon  inators  and  obtain 

U{x+  l)(2a:  -  11)  ^  9  (a:  +  l)(2a:  -  11) 
a;+  1  2a;-  11 

Or,  ll(2x-ll)=9(a:+l).  (2) 

From  the  integral  equation  (2)  we  may  obtain  the  equivalent  integral 

equation  22  x  -  121  =  9  a:  +  9,  (3) 

the  single  solution  of  which  is  found  to  be  a;  =  10. 

Since  neither  of  the  solutions,  a:  =:  —  1  or  a;  =  11/2,  of  the  equation  formed 
by  placing  the  multiplier  {x  -f-  l)(2a;  —  11)  equal  to  zero,  is  a  solution  of 
the  derived  integral  ec] nation  (3),  it  may  be  seen  that  the  single  solution 
x  =  10  of  the  derived  integral  equation  (3)  must  be  the  solution  of  the 
original  fractional  equation  (1),  and  there  can  be  no  other  solution. 

The  solution  may  be  verified  by  substituting  10  for  x  in  (1),  when  we 
shall  obtain  the  identity  1  =  1. 

11  9 

We  may  obtain  the  graph  of  the  equation —  = —  as  follows : 

Transpose  9/(2  x  — 11)  to  the  first   member   and   then  write   the  first 
member  of  the  equation  thus  formed  equal  to  y,  as  follows  : 
11  9 


x+1      2X-11 


=  y.  (4) 


FRACTIONAL  EQUATIONS 


301 


DiflFerent  pairs  of  corresponding  values  of  x  and  y,  which  may  be  taken 
as  abscissas  and  ordinates  respectively  of  points  on  the  graph,  may  be 
obtained  as  follows  : 

By  assigning  different  values  to  x  and  substituting  these  values  for  x  in 
(4)  we  may  calculate  corresponding  values  for  y. 

It  will  be  found  convenient  to  transform  the  first  member  of  equation 
(4)  by  addition,  and  to  obtain  the  values  of  i/  by  substituting  values  of  x 
in  the  transformed  equation  (5). 

13  (a:  -  10) 


-y- 


(5) 


(a:+ l)(2a;-ll) 
After    having    computed    pairs 
of  values  for  x  and  i/,  points  on 
the  graph  may  be  located.     (See 
Fig.  1.) 

7.  If  a  fractional  term  of  an 
equation  is  preceded  by  a  neg- 
ative sign,  then  when  deriving 
an  integral  equation  (by  mul- 
tiplying every  term  of  the 
equation  by  the  lowest  common 
multiple  of  the  denominators 
of  the  different  terms)  it  is 
necessary  to  change  the  sign  of 
every  term  of  the  numerator 
from  -f  to  — ,  or  from  —  to  +. 

X,      «      oti  a:  —  2      a:4-2      x  —\      ^ 

Ex.  2.     Solve  — -— ^  -  -. — -  =  0. 

X  -\-  1      x  —  2      x^  —  ^ 

Clearing  of  fractions  by  nmltiplying  every  term  by  x'^ 

lowest  common  denominator  of  the  fractions,  and  changing  the  signs  of  the 

terms  in  the  numerator  a;  —  1  of  the  last  fraction,  we  have 

(a;  _  2)2  -  (x  +  2)2  -  X  +  1  =  0 


4,  which  is  the 


x^-4x-\-4 


4x-4-a;-l-  1  =0 
-9x  =  - 


By  substituting  1/9  for  x  in  the  given  equation,  the  solution  x  =  1/9 

may  be  verified  as  follows : 

-  17      19       72  _ 
19  "^  17      323  ~ 


289  -f  361 


72  =  0 
0  =  0. 


802  FIRST  COURSE  IN  ALGEBRA 

8.  In  some  cases,  before  clearing  of  fractions,  it  is  well  to  unite 
some  of  the  fractions  appearing  in  an  equation. 

Ex.3.   Solve  — ^ !_  =  _L^ !_.  (1) 

x  —  1      X  — 3       a:  —  5       x—\  ^' 

Uniting  fractional  terms  in  each  member  separately, 

(a:-3)-(x-7)  _  (x-l)-(x-5) 

{x  -l){x-  3)  (X  _  5)(a;  -  1)  '  ^^ 

Simplifying  the  numerators,  we  have 

4  4 

(x  -  7)(x  -  3)  ~  (x  -  b){x  -  1)'  ^^^ 

Instead  of  obtaining  an  integral  equation  by  dividing  both  members  by  4 

and  clearing  of  fractions,  we  may  proceed  as  follows : 

Since  the   members   of  the   equation  are  equal  fractions  having  equal 

numerators,  we  may  ec^uate  the  denominators  at  once,  and  write 

(x  -  7)(x  _  3)  =  (x  -  5)(x  -  1). 

From  this  equation  we  obtain 

x^  -  lOx  +  21  =  x2  -  6x  +  5. 

Collecting  terms,  —  4  x  =  —  16. 

Finally,  we  obtain  x  =  4. 

Since  the  root  4  could  not  have  been  introduced  when  (3)  was  being 

cleared  of  fractions,  it  must  be  the  single  solution  of  the  original  equation  (1). 

The  solution  may  be  verified  by  substituting  4  for  x  in  the   original 

equation. 

General  Directions  for  Solving  Fractional  Equations 

9.  Although  special  devices  may  be  employed  to  obtain  the  solu- 
tions of  fractional  equations,  the  following  general  directions  will 
often  enable  the  student  to  avoid  unnecessary  work,  and  also  to 
avoid  introducing  into  the  derived  integral  equations  roots  which  do 
not  satisfy  the  given  fractional  equations. 

1.  Before  clearing  of  fractions,  all  fractions  should  be  reduced  to 
lowest  terms. 

2.  Fractions  having  a  common  denominator  should  be  combined. 

3.  Wherever  possible,  the  denominators  of  fractions  in  lowest 
terms  should  be  factored  and  the  factored  forms  retained  until  an 
integral  equation  is  derived,  for  the  forms  of  these  factors  may  sug- 
gest a  simple  grouping  of  the  terms  of  the  equation. 

4.  When  clearing  of  fractions,  use  the  lowest  common  multiple 
of  the  denominators  of  the  fractions  in  lowest  terms  as  a  multiplier. 


FRACTIONAL  EQUATIONS  303 

Exercise  XVI.     1 

Solve  the  following  fractional  equations  for  the  letters  appearing 
in  them  -,  the  first  one  hundred  and  twenty  equations  may  be  solved 
mentally : 


1. 

X 

15. 

A='- 

29. 

1-15  =  0. 

a 

2. 

y 

16. 

■=f. 

30. 

1-6  =  0. 

X 

3. 

z 

17.' 

■=si- 

31. 

1-5  =  0. 

y 

4. 

1=1. 

w 

18. 

I'- 

32. 

1-8  =  0. 

Z 

5. 

A  =  -i. 

m 

19. 

A-- 

33. 

14  —  i  =  0. 

a 

6. 

-*  =  i. 

n 

20. 

8-    ■^    . 

34. 

.,-1  =  0. 

7. 

1  =  2. 

X 

21. 

-A- 

35. 

?-8=0. 

c 

8. 

1  =  3. 

y 

22. 

-£■ 

36. 

1  +  5  =  0. 

9. 

4  =  1. 

z 

23. 

-A- 

37. 

1+13  =  0. 

10. 

6  =  1. 

w 

24. 

■»=s- 

38. 

1-2  =  3. 

X 

11. 

h'- 

25. 

5-,.o. 

39. 

1-4  =  7. 

y 

12. 

i."^ 

26. 

1^-1  =  0. 

2/ 

40. 

1-1  =  1. 

Z 

13. 

A=-- 

27. 

11-1=0. 

Z 

41. 

i  +  3  =  6. 

14. 

/.=■• 

28. 

1-1^  =  0. 

t<7 

42. 

l.,  =  s. 

34                        FIRST  COUl^E  IN 

ALGEBRA 

43.  i  +  8  =  2. 
c 

59.  1  =  ^  ^     • 

75. 

1              1 
2a+l      a+3 

44.  5  +  -  =  8. 
w 

««-2.+  l-l- 

76. 

1                1 

36  —  2      26  +  2 

45.  9  +  -  =  7. 
a 

«^-3Al=-- 

77. 

2               2 
5c-l      3c+5 

46.  10  +  7  =  6. 
b 

4 

78. 

1       2 
a:      3* 

47.  4  -  -  =  3. 

X 

«^-.  +  4       ^• 

79. 

1       6 

48.  5  -  -  =  2. 

«*-,  +  2-^- 

80. 

1  _      8 
z           9* 

49.  6  -  -  =  10. 

z 

«^-.+  2-^- 

81. 

4_  1 
7       x' 

50.  7-i  =  I2. 
w 

^^-.-a-'^- 

82. 

3       2 

8~y' 

"■^.='- 

67.        %  =  7. 
m  —  0 

83. 

1       5 

5~2;* 

-jfa- 

''■1=1- 

n       6 

84. 

1         3 
5       2w 

53.       ^  ,  =  1. 
2  —  3 

^'■H- 

85. 

2    _5^ 
7«      9' 

54.  — i— =  1. 

W7—  6 

™.i=-i. 

86. 

3            7 
76~      3' 

55.^1^-1. 

'■J4- 

87. 

4_       7 
5~       6c* 

^«-  3-6-1- 

-.-i-.=r 

88. 

57.  5  =  ^4 — 

0  +  c 

"■F^.  =  l- 

89. 

1/    y 

5°    1-      '' 

".-i-.  =  r 

90. 

?-?-4. 

^-  ^-d-7 

2;         r 

FRACTIONAL  EQUATIONS  305 

9xJ-5  =  2.         101.^  =  3.  111.       2  1 

92.  1+1  =  10-?.  102.  ^-ti  =  4.  112. 
0                    b             a  —  2 

4  3  ^  —  1 

93.  -+3  =  -  +  4.     103.  -r—— =  5.  113. 
c            c                      0  +  6 

94.2_£+|  =  i.       104.  JL  +  Us.         114. 
a;  +  4  2x      X 

96.^^I1«=,.      i06.J-+J-  =  l.      116.       ^  1 


X 

-3 

a;-2 

7 

6 

X 

+  3 

x-2 

5 
—  1 

4 

X 

a:+l 

3 

1 

2 

x+i 

x-l 

7 

2 

m  Sx      2x  bx—1     2ic— 1 

97.  t^Lll^  =  3.       ,07.  ^-A=2.      117.       '  ' 


2y       4y  3a;+2     5a;— 2 

5^Il9_o        108.  a +1=3.      118.^— A- 


2^  bz       2z  3a;  +  4     2a;— 1 

99.:r^  =  2.      109.  4^  = -U-   119.      "  ^ 


3A  +  4  a;+5      a;  +  2  7a;+3     6a;+2 

'''•  41^5  =  ''       '''•  ^  =  d^-   ''''  2-^3-4^ 
121    ^+1       ^-^ 


122. 
123. 
124. 
125. 


a;  —  2      a;  +  5 

4  a;— 3  _  4a;—  7 
2ic—  1  ~2a;  — 5* 

6a;— 2_  2a;+  1 
3a;  +  4""     a;4-  3* 

1  ^  1 

(x  +  l)(a;  +  4)  ~"  (a;  +  2)(a;  +  5) 

1  1 


(a;  +  14)(a;  -  7)       (x  -  13)(a;  -  6) 


12  4a; +30 

126.  3 — T  =  '^ J^V- 

x-\-  1  x+  S 

20 


806  FIRST  COURSE  IN  ALGEBRA 


127. 

1 

1 

2 

3 

X 

— 

2 

1 

X 

-3 

X 

—  4 

128. 

X 

J_ 

"7 

— 

X 

3 
-3 

~~  X 

4 
-4 

129. 

X 

1 

"2 

— 

1 

X 

1 
—  4 

X 

1 

X 

—  1 

3 

130. 

1 

1 

.^. 

1 

1 

«  — 9      jc— 11       x—\h      a— 17 

Equations  in  which  Numbers  other  than  the  Unknown  are 
Represented  by  Letters 

10.  Equations  in  which  coefficients  and  known  numbers  are 
represented  by  letters  are  called  literal  ecj nations. 

In  the  preceding  chapters  principles  were  developed  which  may 
be  applied  to  obtain  the  solutions  of  literal  equations  containing 
ing  one  unknown  number. 

11.  A  literal  equation  is  said  to  be  solved  with  reference  to  a 
specified  letter  when  the  value  of  this  letter  is  expressed  in  terms  of 
the  remaining  letters  appearing  in  the  equation.  We  shall  com- 
monly refer  to  the  value  thus  obtained  as  the  expressed  value 
of  the  unknown. 

It  should  be  understood  that  no  numerical  value  is  obtained  for 
the  unknown  by  this  process. 

Numerical  values  for  the  specified  unknown  letters  can  be  ob- 
tained only  when  definite  values  are  assigned  to  the  letters  which 
are  regarded  as  representing  known  values. 

Liiteral  Equations  wliicli  are  Integral  with 
Reference  to  the  Unknown 

Ex.  1.    Regarding  x  as  the  unknown,  solve  x  -^  d  —  c. 
Transposing,  we  have,  x  —  c  —  d. 

This  expressed  value  is  found  by  substitution  to  satisfy  the  given  equation. 
Ex.  2.   Regarding  x  as  the  unknown,  solve  ax  -\-n  —  m. 
We  have,  ax  =  m  —  n. 

Therefore  x  = 


a 
This  expressed  value  will  be  found  to  satisfy  the  given  equation. 


LITERAL  EQUATIONS  307 

Mental  Exercise  XVI.     2 

Regarding  a;,  2/,  ^,  and  w  as  unknowns,  solve  the  following  literal 
equations  which  are  integral  with  reference  to  x,  y,  z^  and  w  : 

\.  x^h  =  a,  34.  ab'^cx  =  a%c^, 

2.  a;  +  6?  =  c. 

3.  n  =^  m  —  z. 

4.  k  +  /i  =  \o, 

5.  a;  +  1  =  c. 

6.  2/  4-  4  =  ^. 

7.  2;  —  G  3=  ^. 

8.  IV -0  =  7. 

9.  aa;  =  6. 
10.  by  =  c. 
IL  az  =  -h. 

12.  «  =  ^. 

13.  —  c  =  dx. 

14.  ax  —  b  =  1. 

15.  aic  4-  c  =  1. 

16.  bij  —  2  =  c. 

17.  S  +  mz  =  n. 

18.  aa;+  ^  =  c. 

19.  by  —  c  =  a. 

20.  C2;  —  6  =  —  a. 

21.  aa:  +  ^  =  «c. 

22.  a  =  ^a;  +  c. 

23.  r  =  sx  —  t. 

24.  m  =  n  —  qy. 

25.  ace  =  6  +  c. 

26.  2  a^ic  =  a  +  6. 

27.  a^a;  =  Of  +  1. 

28.  (m  +  n)x  —  2  mn. 

29.  («  +  5)  ?/  =  5  a. 

30.  (3  —m)z  =  3  w. 

31.  aa;  =  a\ 

32.  ^/^a;  =  b. 

33.  «fca;  =  6c. 


35. 

— 

abx  =  a  - 

-b. 

36. 

(a  -\-  b)x  =  c 

'■^d. 

37. 

(c 

-2)y  =  c 

^  +  3. 

38. 

(a 

+  b)y  =  a 

f^  -  b"". 

39. 

(m  —  w)  z  = 

m^  —  n 

40. 

(a 

--4)a.= 

a -2. 

41. 

(b 

'-\)y  = 

b+h 

42. 

x 
a 

=  b. 

43. 

y 
b 

=  c. 

44. 

r 

+  s 

45. 

m 

— —  =  mn 

—  n 

46. 

a 

l,^  =  ,a. 

47. 

a 

X                  3 

48. 

X 

c 

1 
~  b' 

49. 

X 

a 

_b^ 

~  c 

50. 

X 

c 

c 

~~d 

51. 

X 

2 

2 

a 

52. 

y . 
b 

_b 

308  FIRST  COURSE   IN  ALGEBRA 


71.  by  —  h  =  nij  —  n. 

72.  mx -V  r^  =^  nx  •\-  m\ 
7.3.  ax  =  bx-{-  1. 

74.  dy  =  \  —  ny. 

75.  2a}j  =  '6by-\-  4. 

76.  abx  +  bcx  =  ca. 

77.  aaj  —  to  —  1  =  cjB. 

78.  ay  +  ^//  =  d  —  cy. 

79.  a{y  —  b)  =  c. 

80.  a(ic+  1)=^. 

81.  c(%  +  1)  =  c?. 

82.  b{b  —y)=a. 

83.  a(l  —  bx)  =  b. 


84.  -  -  1  =  6. 


85.  ^  +  1  =  c. 
a 


86.  -4-7  -b  =  a. 
a  -\-  b 

^„    (IX  ,  bx 

87.  —  +  —  =  2. 

«>        a 


12.  Literal  Equations  as  Formulas.  Instead  of  stating  a 
mathematical  law  or  principle  in  words,  it  is  often  more  convenient 
to  express  it  by  means  of  a  literal  equation  in  which  the  letters 
used  are  understood  as  standing  for  particular  quantities. 

When  so  used,  a  literal  equation  is  called  a  formula. 

In  applications  of  mathematics  to  other  sciences,  it  is  a  common 
practice  to  state  principles  by  means  of  formulas. 

E.  g.  The  identity  (a  ±  hy  =  a^  ±  2  a6  +  h'^  is  a  formula  for  finding  the 
square  of  a  binomial  sum  or  difference. 

The  literal  equation  i=zp  x  r  x  t  may  be  used  as  a  formula  for  comput- 
ing simple  interest,  provided  that  the  lettei-s  i,  p,  r,  and  t  are  understood  aa 
representing  interest,  principle,  rate,  and  time  respectively. 


53. 

n 

=  1. 

54. 

dw 
T 

=  — 

1. 

55. 

to 
c 

1 
~  d' 

56. 

ax 

T 

_b 
~  a 

57. 

X  — 

■  m  = 

n  - 

-  » 

58. 

a  — 

■  x  = 

X  — 

a. 

59. 

b- 

x  = 

X  — 

c. 

60. 

dy- 

-  1  = 

=  1  - 

-dy. 

61. 

ax  • 

-b  = 

--b- 

-ax. 

62. 

hz- 

-3  = 

=  3- 

-hz. 

63. 

a  — 

■bx  = 

--bx 

—  a. 

64. 

b- 

ex  = 

■  ex  ■ 

-b. 

65. 

1  - 

•  ax  = 

=  ax 

—  1. 

66. 

2- 

cy  = 

■cy- 

-2. 

67. 

ax-\-  bx 

=  a 

+  b. 

68. 

h-\- 

mx- 

=  to  +  m. 

69. 

to- 

-  ex  - 

=  b 

—  c. 

70. 

Jcy 

-k  = 

--ry 

—  r. 

m  ^ 


LITERAL  EQUATIONS  309 

13.  When  solving  numerical  equations,  certain  terms  are  often 
so  combined  as  to  cause  particular  numerical  constants  to  disappear 
from  the  calculation ;  but,  when  solving  literal  equations  or  for- 
mulas, it  often  happens  that  none  of  the  given  letters  disappear,  but 
may  be  traced  throughout  the  entire  work. 

14.  The  solution  of  a  particular  literal  equation  may  be  used  to 
obtain  the  solutions  of  an  indefinite  number  of  numerical  equations 
of  corresponding  typey  that  is,  of  equations  in  which  numerical  con- 
stants appear  in  place  of  the  literal  constants  of  the  given  literal 
equation. 

The  particular  literal  equation  thus  solved  may  be  used  as  a 
formula  for  obtaining  the  solutions  of  the  numerical  equations  which 
it  may  be  taken  as  representing. 

E.  g.  Let  it  be  required  to  solve  several  ec^uations  such  as  the  following : 

6xH-    4  =  5a;-f    7,  (1) 

Sx+    3  =  2x4-11,  (2) 

2x-\-    9=    ar-M4,  (3) 

'Sx-l0  =  4x+\8.  (4) 

All  of  the  equations  are  of  the  type  az  +  b  =  ex -\-  d,  in  which  a,  h,  c,  and  d 
represent  in  the  first  equation  6,  4,  5,  and  7  respectively ;  in  the  second 
equation  8,  3,  2,  and  11  respectively;  etc. 

We  may  obtain  the  solution  of  the  literal  equation  as  follows : 

From  ax  -\- h  =  ex  -\-  d, 

we  obtain  ax  —  ex  =  d  —  h. 

Hence,  x(a  —  c)  =  d  —  b, 

d-b 
and  finally,  ^^^TT^' 

From  the  process  of  derivation  the  successive  equations  are  equivalent, 
and  hence  solutions  have  neither  been  gained  nor  lost. 

This  solution  may  be  verified  by  direct  substitution. 

The  solution  x  =  (d  -  b)/(a  -  c),  which  is  a  literal  equation,  may  be  used 
as  a  formula  for  obtaining  numerical  values  for  x  in  the  given  equations  by 
substituting  for  a,  &,  c,  and  d  the  values  which  they  represent. 

7-4 


For  equation  (1),  we  have,  x  = 


=  3. 


11-3      4 
For  equation  (2),  we  have,  x  =    q  _2  ~  3' 


310  FIRST  COURSE  IN  ALGEBRA 

14  —  9 

For  equation  (3),  we  have,  x  =  — =  5. 

^  —  i 

T^  ..  ,4X  1  18    +    10 

For  equation  (4),  we  bave,  x  =  — =  —  28. 

Ex.  1.     Solve  a  (x  -  a)  -  2ab  =  b  {b  -  x).  (1) 

Removing  parentheses  by  performing  the  indicated  multiplications,  we 
obtain  by  Principle  I,  Chap.  X.  §  25,  the  equivalent  equation 

ax-a^-2ab=:b^-bx.  (2) 

Transposing  to  the  first  member  the  terms  containing  x,  and  to  the  second 
member  those  free  from  x,  we  obtain  by  Chap.  X.  §  27,  the  equivalent 
equation 

ax-^bx  =  h^  +  a^  +  2ab.  (3) 

Factoring  the  members  separately,  we  obtain  by  Principle  I,  Chap.  X. 
§  25,  the  equivalent  equation 

(a  +  b)x=(a-\-by.  (4) 

Therefore  x  =  ^^44^  =  a  +  b, 

a  +  0 

which  by  the  process  of  derivation  must  satisfy  the  original  equation  and  be 

its  only  solution. 

We  may  verify  this  result  by  substituting  in  equation  (1)  as  follows: 

a(a  +  b  —  a)  —  2  ab  =  6[6  —  (a  +  A)] 

a(b)-2ab  =  b(b-a-b) 

ab-2ab  =  Z>(-  a) 

—  ab  =  —  ab. 

Numerical  Checks  for  the  Solutions  of  Literal  Equations 

15.  Whenever  numerical  values  are  assigned  to  the  letters  repre- 
senting known  quantities  in  a  literal  equation,  a  numerical  equation 
is  obtained.  The  solutions  of  this  numerical  equation  are  equal 
to  the  numerical  values  found  by  substituting  the  same  numerical 
values  for  the  known  letters  in  the  expressed  value  for  the  unknown 
which  is  the  solution  of  the  given  literal  equation. 

It  follows  that  the  values  assigned  to  the  known  letters  and  the 
value  calculated  for  the  unknown  letter  will,  if  substituted  for  the 
known  and  unknown  letters  respectively  in  the  given  literal  equa- 
tion, reduce  it  to  a  numerical  identity. 

Regarding  x  as  the  unknown,  we  obtain  from  the  literal  equa- 


LITERAL  EQUATIONS  811 

tion  a{x  —  a)  —  2ab  =  b(b  —  x),  the  expressed  value  x  =  a  +  b. 
(See  Ex.  1.  §  14.) 

By  assigning  particular  numerical  values  to  a  and  b  we  may, 
from  the  expressed  value  x  =  a  +  b,  calculate  a  numerical  value 
for  X. 

Thus,  if  a  =  I,  and  b  =  2,  it  follows  from  x  =  a  +  b  that  x  =  S. 

Substituting  1  for  a,  2  for  b  and  3  for  x  in  the  given  literal  equa- 
tion, a(x  —  a)  —  2  ab  =  b(b  —  x),  we  obtain  the  numerical  identity, 
1  (3  —  1)  —  2  •  1  •  2  =  2  (2  —  3),  which  reduces  to  —  2  =  —  2. 

Accordingly,  we  have  by  this  method  verified  for  particular 
values  of  the  letters  the  solution  a;  =  «  +  ^  of  the  given  literal 
equation. 

From  the  nature  of  the  method  it  may  be  seen  that  if  the  check 
holds  in  a  particular  case  it  holds  in  all  cases. 

Hence  we  have  verified  the  solution  of  the  given  literal  equation. 

Exercise  XVI.    3 

Solve  the  following  literal  equations  for  x,  verifying  all  results 
either  by  substituting  the  literal  solutions  directly,  or  by  making 
proper  numerical  substitutions  : 

1.  a(x  •\-b)  =  b(x  +  a). 

2.  ax-\-  bc  =  d(b  -\-  x). 

3.  {x  +  a)(x  +  b)  =  x{x  -f  c). 

4.  m  =  a  -{-  {n  —  \)x. 

5.  {m  —  n)x  —  m^  =  (m  +  n)x. 

6.  a(l  +  x)  +  6(1  +x)=x(a  +  b  +  1). 

7.  a(x  —  1)  +  («  —  l)x  =  a  +  x. 

8.  ala  —  2x)  +  b(b  —  2x)  +  2ab  =  0. 

9.  3(3 x-b)  +  2b  =  b(bx  —  3)  +  6. 

10.  hk{x^  -  1)  =  (^  +  kx){k  +  hx). 

11.  (x  +  a){x  -b)  =  {x  +  a-  by. 

12.  (b  -  c)(x  -'b)  =  (b-  d)x. 

13;  \x  -  a){b  -  c)  =  (a  -  c)(x  -  b). 

14.  {a  +  by  +  («  -  x)(b  -x)  =  {x  +  d){x  +  b). 

15.  \a  -  x){x  +  b)-  c(a  +  c)  =  (c  -  x)(x  +  c)  +  ab. 

16.  -^  +  b  =  a. 
a  +  b 


812  FIRST  COURSE  IN  ALGEBRA 

a        0 

19.^  +  1^=1. 
a         c 

X        .         X 

20.  ~-\-b  =  j  +  a, 
a  0 


21. 

a; 
9 

-h 

X 

-9- 

22. 

X 

m 

+  ^ 

—  X 

n 

=  6. 

23. 

X 

a 

m 

—  X 

b 

=  n. 

4U. 

6      ""     a 

27. 

d  +  X          X 

k     ^  d-\'k 

28. 

x  —  a  ^  X  —  b      a^  +  b^ 
b       ^      a     ~     ab 

29. 

b^  —  ax      ,       a^  —  bx 
b       =''          a      ■ 

30. 

b             a      a      b 

31. 

ex 

=  c  —  a. 

c  —  d 

32. 

c^x 

bx      «^-^_^- 

33. 

a-\-  X      (ix  _b 
a  -f  6       b       a 

34. 

CMC      bx      ex        - 
be      ae      ab       ' 

24.f  =  .-6  +  J. 

25,  bx  +  b  =  ^-\-l' 
b      b 

Literal  Equations  which  are  Fractional  with 
Reference  to  tlie  Unknown 

16.   When  solving  literal  equations  which  are  fractional  with 

reference  to  certain  specified  letters  it  is  often  convenient  to  obtain 

the  solutions  by  deriving  integral  equations  in  which  the  unknown 

letters  are  found  in  the  second  member  instead  of  in  the  first. 

Ill 
Ex.  1.     Solve  for  x,  —  =  n. 

X 

Clearing  of  fractions,  we  have 

m  =  nx. 

Hence,  —  =  x. 

n 

Ex.  2.    Solve  for  «,  i—  =  c. 

X  —  t> 

Clearing  of  fractions  we  obtain 

ab  —  be  =  ex  —  he. 

From  which,  ab  =  ex. 

Hence,  —  =  x. 

c 


LITERAL  EQUATIONS  313 

Mental  Exercise  XVI.   4 
Solve  the  following  equations  for  a,  y,  z^  and  w  : 

m  —  n 


1. 

a 

X 

b  _ 

=  1. 

2. 

y~ 

=  1. 

_d 

3. 

1  = 

z 

4. 

1  = 

1 

m 

=  — ' 
w 

5. 

2 

=  m. 

6. 

1)' 
4 

=  n. 

7. 

z 
7 

=  k. 

8. 

w  ' 

=  q. 

9. 

c 

X 

k  _ 

=  2. 

10. 

y~ 

=  3. 
r 

11. 

4  = 

~  z 

12. 

8  = 
h 

t  ^ 
'  w 

13. 

X 

he 

-  c. 

14. 

X 

—  a. 

15. 

d  _ 

y~ 

■fif' 

16. 

a  _ 

X 

2 
3* 

17. 

b_ 

y 

5 
9* 

18. 

m  _ 
X  " 

=  -  1. 

19. 

n 

y~ 

-2. 

20. 

9  = 

_b 

y 

21. 

ax 

=  1. 

22. 

2 

=  1. 

23. 

3  _ 

C1J~ 

=  —  1. 

24. 

4  = 

_3_ 

mz 

25. 

mn 

X 

=  q. 

26. 

r  __ 
sx" 

:t. 

27. 

a-\-  b              , 
=  a  —  b. 

X 


28.  ^-^=c  +  d. 

y 


A  t/l 

1       lllr 

-r  II'  — 

z 

30. 

,      k' 

-1  =  ^+1. 

w 

31. 

1 

X 

a 

32. 

d 
n 

__  1 

X 

33. 

g 

n 

_  1 

y' 

34. 

X 

c 

~    ~d 

35. 

3 

X 

4 

a 

36. 

5 

y' 

b 
6* 

37. 

m 
2 

3 

~  z 

38. 

a 

X  ' 

_b  ^ 
a 

39. 

5 

_h 

z 

40. 

b  _ 
c 

_  c 

~y' 

41. 

a 

2^ 

2 

ax 

42. 

b 
3 

3 

314 


FIRST  COURSE   IN  ALGEBRA 


43.«=£- 
CZ      6 

58. 

44     «   =1 
rfa-      8 

59. 

45.^  =  1. 
6      ax 

60. 

46.^  =  ^- 

61. 

47    2«^   c  . 

62. 

Aft 

^      -1 

*o. 

w  —  I 

49. 

\-l 

y-l 

50. 

1       _1 

X  —  a 

51. 

1          1 

iC  —  c  ~  6 

52. 

1        1 

y  -b~  c 

53. 

1           1 

z  +  d      m 

54. 

1           1 

55. 

1           1 

a      a;  —  a 

56. 

1           1 

b     y-b 

^1 

1            1 

J J^ 

-\-2b~  bb 
1  1 


73.  -r  =  3. 

y-b 


X  —  a      a  —  b 

X  -^^  c      c  -{-  d 
1  1 


z  +  a      Sa 


63. 
64. 
65. 
66. 
67. 
68. 
69. 
70. 
71. 
72. 


z-g 

g-h 

1 

1 

2x  —  m 

m  —  2 

1 

1 

Sy  —  t 

^-3 

1 

1 

X  —  a 

'6  +  c 

1 

1 

x+  b~ 

a  +  c 

1 

1 

x  +  c 

a  +  6 

1 

1 

ax  +  b' 

~6  +  c 

a 

1. 

x-b~ 

c 

1. 

y  +  d" 

2k 

z  +  h 

1. 

Am 

—  1 

w  +  m 

a 

:2. 

74. 


75. 


76. 


77. 


78. 


79. 


80. 


z-h 


z\d 


=  4. 


=  2. 


=  3. 


=  4. 


y-3 


z^-h 


=  b. 


=  L 


81. 

w+1-^'' 

82. 

5-      ^      . 

^-^-2 

83. 

,',-■ 

84. 

.-^■=- 

85. 

a 

r  =  C. 

CC—  ^ 

86. 

m 
=  n. 

y  —  m 

87. 

,  "   -(^. 

X  —  a 


bx  -\-  c 


LITERAL  EQUATIONS  315 

„            „  ^      ax  —  h      hx  —  a        ,  „ . 

Ex.  1.    Solve     ^^     =  —^ ah.  (I) 

Clearing  of  fractions  by  multiplying  each  of  the  terms  by  the  lowest 
common  denominator,  abx^  of  the  fractional  terms,  we  obtain 

ahx  -b^  =  abx  -  a^  -  a%'^x.  (2) 

Hence,                                  a%'^x  =  b^-  a^.  (3) 

m,,      .  &'  - «' 

Therefore,  x  =  — ^—- . 

We  may  apply  the  metliod  of  §  15  and  verify  this  solution  as  follows: 

9  —  4 
If  we  let  a  =  2  and  &  =  3,  it  follows  that  x  =  — — —  =  ,\. 

Substituting  2,  3,  and  5/36  for  a,  &,  and  x  respectively,  in  the  given  literal 
equation,  we  obtain  the  following  numerical  equality  which  is  found  to  be 
a  numerical  identity  : 


ii)  ~  KA) 


-2-3 


__49__49 
5  ~       5  * 

Exercise  XVI.    5 
Regarding  x  as  the  unknown,  solve  the  following  literal  equations  : 
h      g      1 

^  X         X 

2.  '^  =  m{n-t)  +  -' 

XX 


4:.     b   = 


c  +  dx 
c  —  X 
ax 


e  —  a 
b.  n  = h  1. 

X 


e.  m  = 


7.  ^*-  =  l. 

dx  +  kx 


8. 

X  —  a 

X  —  s 

9. 

x+b      3 
x-b      4 

10. 

X  —  a      a^ 
x-b~b^ 

IL 

b       x-a^ 
a      x  —  b'^ 

12. 

c  —  dx      c 
ex  —  d      d 

13. 

c  —  ax              ( 

14 

X          a  -\-  b 

a  +  K^-m)  ,,   _^:^  +  i  +  ^  =  0 


bx 
a 


316  FIRST  COURSE   IN  ALGEBRA 

15.2j!i*  =  _^.  19. 


g-x 

-g-h 

x  +  p  _ 

p-q 

x-q 

p  +  q 

e  —  X 

d-x 

1%.-^^-^-^=^^--^'  20. 


17.  -,= 21. 


X 

— 

0 

X 

— 

e 

X 

— 

4 

X 

— 

5 

a 

— 

X 

= 

X 

a 

+ 

X 

a 

— 

X 

X 

— 

a 

8 

X  +  a 


x—d      x—e  '       3  X + a  3 

^^rn-^^m-X  22.1  +  1  +  1  +  1  =  1. 

n—xn—\  a      0      c      a      X 

-      *     26c  2ac  ^   2ab       a      b      c 


2x  +  b 

Exercise  XVI.     6 

Problems  Solved  by  Fractional  Equations 

Solve  the  following  problems,  examining  the  solutions  to  see  if  they 
satisfy  the  conditions  of  the  problems  as  stated : 

1.  What  must  be  the  value  of  a  in  order  that  (3a4-  10) /(14a  — 4) 
shall  have  the  value  1  /2  ? 

2.  Find  two  numbers,  whose  sum  is  73,  which  are  such  that  the  quotient 
obtained  by  dividing  the  greater  by  the  less  is  3  and  the  remainder  13. 

If  X  represents  the  greater  number,   73  —  a:   will   represent  the   less 
number. 

By  the  conditions  of  the  problem,  we  have  the  conditional  equation 

'      =3+      '^ 


73 -a;  '  73-a; 

from  which  x  is  found  to  be  58,  which  is  the  greater  number. 

Accordingly  the  less  number,  which  is  represented  by  73  —  x,  is  15. 
These  numbers  are  found  to  satisfy  the  conditions  of  the  problem. 

3.  Find  two  numbers,  whose  sum  is  .36,  which  are  such  that  the  quotient 
obtained  by  dividing  the  less  by  the  greater  is  2/7. 

4.  The  sum  of  two  numbers  is  28,  and  the  quotient  obtained  by  dividing 
the  less  by  the  greater  is  3  /  4.     Find  the  numbers. 

5.  The  sum  of  two  numbers  is  59,  and  if  the  greater  be  divided  by  the 
less,  the  quotient  is  3  and  the  remainder  is  3.     Find  the  numbers. 


PROBLEMS  S17 

6.  The  sum  of  two  numbers  is  116,  and  if  the  greater  be  divided  by  the 
less  the  quotient  is  8  and  the  remainder  8.     Find  the  numbers. 

7.  The  difference  between  two  numbers  is  60 ;  if  the  greater  is  divided 
by  the  less  the  quotient  is  7  and  the  remainder  6.     Find  the  numbers. 

8.  What  number  must  be  added  to  the  numerator  and  also  to  the 
denominator  of  the  fraction  41/57  in  order  that  the  resulting  fraction  shall 
equal  11/15? 

9.  What  number  must  be  added  to  the  numerator  and  subtracted  from 
the  denominator  of  the  fraction  7/12  in  order  that  the  result  shall  be  equal 
to  the  reciprocal  of  the  given  fraction  ? 

10.  When  four  is  subtracted  from  the  numerator  of  a  fraction  of  which 
the  numerator  is  three  less  than  its  denominator,  the  value  of  the  fraction 
becomes  one-eighth.     What  is  the  original  fraction  ? 

11.  The  reduced  value  of  a  certain  fraction  is  3/7  and  its  denominator 
exceeds  its  numerator  by  20.     Find  the  fraction. 

12.  The  value  of  a  fraction  is  1/12.  If  its  numerator  is  increased  by  5 
and  its  denominator  by  4,  the  resulting  fraction  will  be  equal  to  1/5.  Find 
the  fraction. 

13.  Separate  580  into  two  parts  such  that  when  the  greater  part  is  divided 
by  the  less  the  quotient  is  12  and  the  remainder  is  21. 

14.  The  recii)rocal  of  a  number  is  equal  to  four  times  the  reciprocal  of 
the  sum  of  the  number  and  18.     Find  the  number. 

15.  The  figure  in  units'  place  of.  a  number  expressed  by  two  figures 
exceeds  the  figure  in  tens'  place  by  4.  If  the  number,  increased  by  11,  is 
divided  by  the  sum  of  the  figures  in  units'  and  tens'  places,  the  quotient 
is  5.     What  is  the  number? 

16.  The  figure  in  tens'  place  of  a  number  of  two  figures  exceeds  the  figure 
in  units'  place  by  5,  and  if  the  number  increased  by  5  is  divided  by  the  sum 
of  its  figures  the  quotient  is  8.     Find  the  number. 

17.  A  can  do  a  piece  of  work  in  16  days,  and  B  in  12  days.  How  many 
days  will  be  required  if  both  work  together  ? 

If  X  represents  the  number  of  days  which  will  be  required  when  A  and 
B  work  together,  then  \/x  will  represent  the  fractional  part  of  the  work 
which  will  be  performed  in  one  day.  According  to  the  statement  of  the 
problem,  A  in  one  day  can  perform  1/16  of  the  work,  while  B  can  perform 
1/12. 

By  the  conditions  of  the  problem,  we  have 

16  "^12"  a:* 
Hence  ^  -  ^^ 


318  FIRST  COURSE  IN  ALGEBRA 

Accordingly  6^  days  will  be  required  when  both  work  together. 
This  result  will   be  found  to  satisfy  the  conditions  of  the  problem  as 
stated. 

18.  A  and  B  together  can  paint  a  house  in  12  days,  A  and  C  together  in 
16  days,  and  A  alone  in  20  days.  In  what  time  can  B  and  C  together 
paint  it  ?     In  what  time  can  A,  B,  and  C  working  together  paint  it? 

19.  A  can  do  a  piece  of  work  in  12  days,  B  in  15  days,  and  A,  B,  and 
C  together  in  5  days.     In  how  many  days  can  C  do  the  work  ? 

20.  A  can  do  a  piece  of  work  in  10  days,  B  in  8  days,  and  C  in  5  days. 
How  many  days  will  l>e  required  if  all  work  together  ? 

21.  A  can  do  a  certain  piece  of  work  in  2^  days,  B  in  3}  days,  and  C  in 
4^  days.  If  A,  B,  and  C  work  together,  how  long  will  it  take  them  to  do 
the  work? 

22.  A  sum  of  $1200  was  to  be  divided  equally  among  a  certain  number 
of  persons.  If  there  had  been  four  more  persons,  each  would  have  received 
2/3  as  much.     How  many  persons  were  there  ? 

23.  A  cask  may  be  emptied  by  any  one  of  three  taps.  It  can  be  emptied 
by  the  first  alone  in  25  minutes,  by  the  second  alone  in  30  minutes,  and  by 
the  third  alone  in  40  minutes.  What  time  would  be  required  to  empty  the 
cask  by  using  all  three  together  ? 

24.  A  vat  in  a  paper  mill  can  be  filled  by  one  pipe  in  one  and  one-third 
hours,  by  a  second  in  two  and  one-half  hours,  and  by  a  third  in  four  hours. 
What  time  will  be  required  to  fill  it  when  all  are  running  together  ? 

25.  A  train  runs  164  miles  in  a  given  time.  If  it  were  to  run  3  miles 
an  hour  faster  it  would  go  12  miles  farther  in  the  same  time.  Find  the 
train's  rate  of  speed  in  miles  per  hour. 

26.  It  is  observed  that  a  steamer  can  run  60  miles  with  the  current  in 
the  same  time  that  it  can  run  36  miles  against  the  current.  Find  the  rate 
of  the  current  in  miles  per  hour,  knowing  that  the  steamer  can  run  12 
miles  an  hour  in  still  water. 

27.  A  can  row  five  miles  and  B  four  miles  an  hour  in  still  water.  A  is 
12  miles  farther  up  stream  than  B,  and  they  row  toward  each  other  until 
they  meet  4  miles  above  B's  starting-place.  Find  the  rate  of  the  current 
in  miles  per  hour. 

General  Problems 

17.  Since  a  particular  letter  may  represent  more  than  one  value, 
it  is  customary  to  speak  of  numbers  represented  by  letters  as 
literal  or  general  numbers. 

18.  A  general  problem  is   a  problem  in  which  the   numbers 


PROBLEMS  319 

whose  values  are  supposed  to  be  known  are  represented  by  letters ; 
that  is,  in  which  the  known  numbers  are  general  numbers. 

Exercise  XVI.     7 

Problems  Involving^  Literal  ^Equations 

Find  the  general  solution  of  each  of  the  following  problems : 

1.  Separate  the  number  a  into  two  parts  such  that  m  times  the  first 
part  shall  exceed  n  times  the  second  part  by  h. 

If  X  stands  for  one  of  the  required  numbers,  a  —  x  will  represent  the 
other.     By  the  conditions  of  the  given  problem  we  have 
mx  =.n  X  (a  —  a;)  +  ft, 

from  which  we  obtain  x  = • 

m  +  n 

Hence,  one  of  the  parts  into  which  a  is  required  to  be  separated  is 

4.  1  V                                    an-^b 
represented  by  • 

The  other  part,  represented  by  a  —  a:,  may  be  found  as  follows : 

an  +  6      am  —  b 


a  —  x  =  a 


m  -{-  n 


By  giving  particular  values  to  the  letters  appearing  in  the  state- 
ment of  any  general  problem,  it  is  possible  to  obtain  as  many 
separate  special  problems  of  a  given  type  as  may  be  desired.  The 
solution  of  the  general  problem  will  in  every  case  be  the  solution 
of  all  of  the  special  problems  of  the  given  type. 

E.  g.  The  following  is  a  special  problem  which  is  of  the  same  type  as 
Ex.  1  which  is  a  general  problem : 

Separate  the  number  19  into  two  parts  such  that  10  times  the  first  part 
shall  exceed  3  times  the  second  part  by  8. 

The  solutions  of  this  special  problem  may  either  be  obtained  directly  by 
solving  a  conditional  equation,  or  by  substituting  the  values  a  =  19,  6  =  8, 

an  -\-b  , 
m  =  10  and  71  =  3  for  the  letters  appearnig  m  the  expressions  ^  _^  ^  ant^ 

~      which  are  found  by  solving  the  general  problem. 


The  solutions  of  the  special  problem  are  found  to  be  5  and  14. 
2.    Separate  a  into  two  parts  such  that  m  times  the  first  part  shall  equal 
n  times  the  second  part. 


320  FIRST  COURSE  IN  ALGEBRA 

3.  The  sum  of  two  numbers  is  s  and  their  difference  is  d.  What  are  the 
numbers  ? 

4.  Separate  a  into  three  parts  such  that  the  first  part  shall  be  m  times 
the  second  part,  and  the  second  part  n  times  the  tliird  part. 

5.  Sepamte  d  into  two  parts  such  that  when  one  part  is  divided  by  the 
other  the  quotient  shall  be  q  and  the  remainder  r. 

6.  What  number  must  be  added  to  each  term  of  the  fraction  a/b  in 
order  that  the  resulting  fraction  sliall  be  equal  to  c/d. 

7.  A  can  do  a  piece  of  work  in  a  hours,  and  B  can  do  the  same  piece  of 
work  in  b  hours.     How  many  hours  will  be  required  if  both  work  together  ? 

8.  One  pipe  can  fill  a  tank  in  a  houre  and  a  second  pipe  can  fill  it  in  b 
hours.  If  a  thirtl  pipe  can  empty  it  in  c  hours,  how  many  hours  will  be 
required  to  fill  the  tank  when  the  three  pipes  are  open  ? 

Discuss  the  problem  for  a  +  6  =  c. 

9.  A  tank  can  be  filial  from  three  taps.  By  using  the  first  alone  it  is 
filled  in  a  minutes,  by  using  the  second  alone,  in  6  minutes,  and  by  using 
the  third  alone,  in  c  minutes.  In  how  many  minutes  would  it  be  filled  if 
the  taps  were  all  open  at  the  same  time  ? 

10.  In  how  many  years  will  P  dollars  amount  to  A  dollars  at  r  per  cent 
simple  interest  j)er  year  ? 

11.  What  principal  at  r  per  cent  interest  per  year  will  amount  to  A 
dollars  in  t  years  ? 

12.  An  alloy  of  two  metals  is  composed  of  a  parts  of  one  to  b  parts  of 
the  other.  How  many  pounds  of  each  are  required  to  make  c  pounds  of 
the  alloy  ? 

13.  Two  trains,  A  and  B,  d  miles  apart,  start  at  the  same  time  and  travel 
toward  each  other  at  the  rates  of  a  miles  per  hour  and  6  miles  per  hour 
respectively.     How  far  will  each  have  travelled  when  they  meet  ? 

14.  In  a  certain  time  a  train  ran  a  miles.  If  it  had  run  6  miles  an  hour 
faster  it  would  have  gone  c  miles  farther  in  the  same  time.  Find  the  rate 
of  the  train  in  miles  per  hour. 

15.  If  A  and  B  can  travel  at  the  rates  of  a  and  b  miles  an  hour  respec- 
tively, how  far  must  A  travel  to  overtake  B,  if  both  move  in  the  same 
direction,  and  B  be  given  a  start  of  s  miles  ? 

16.  A  naphtha  launch  can  run  a  miles  an  hour  in  still  water.  If  it  can 
run  6  miles  against  the  current  in  the  same  time  that  it  can  run  c  miles 
with  the  current,  what  is  the  rate  of  the  current  in  miles  per  hour  ? 

17.  A  crew  can  row  a  certain  distance  up  a  stream  in  a  hours  and  can 
row  back  again  in  6  hours.  If  the  rate  of  the  crew  in  still  water  is  s  miles 
an  hour,  find  the  velocity  of  the  stream  in  miles  per  hour. 


PROBLEMS  321 

18.  A  dealer  mixes  a  pounds  of  tea  worth  x  cents  a  pound  with  h  pounds 
of  tea  worth  y  cents  a  pound  and  with  c  pounds  of  tea  worth  ^  cents  a  pound. 
Find  the  value,  v,  of  the  mixture  in  cents  per  pound. 

19.  Pieces  of  money  of  one  denomination  are  of  such  value  that  a  pieces 
are  equal  in  value  to  one  dollar,  and  pieces  of  money  of  another  denomina- 
tion are  of  such  value  that  h  pieces  are  equal  in  value  to  one  dollar.  Find 
how  many  pieces  of  each  denomination  must  be  taken  on  condition  that  c 
pieces  of  money  shall  he  equal  in  value  to  one  dollar. 

The  Interpretation  of  the  Solutions  of  Problems 

19.  It  often  happens  that  there  are  restrictions  on  the  nature  of 
the  unknown  numbers  of  a  given  problem  which  cannot  be  trans- 
lated into  algebraic  language,  and  hence  cannot  be  expressed  by 
means  of  algebraic  conditional  equations. 

If,  for  example,  the  unknown  number  of  a  problem  represents  a 
number  of  men,  it  is  implied  that  the  number  sought  is  integral, 
yet  this  implied  condition  cannot  be  translated  into  algebraic  lan- 
guage and  expressed  in  a  conditional  equation. 

20.  It  appears  that,  when  solving  the  conditional  equations  aris- 
ing from  the  translation  into  algebraic  language  of  the  conditions 
of  a  stated  problem,  all  that  we  know  at  the  outset  is  that  the  solu- 
tions of  the  problem,  if  indeed  any  exist,  must  be  found  among  the 
algebraic  solutions  of  the  conditional  equations. 

If  it  happens  that  none  of  these  solutions  are  consistent  with  the 
stated  conditions  of  the  problem,  we  conclude  that  the  concrete 
problem  as  stated  has  no  solution. 

Ex.  1.  At  an  entertainment  75  cents  was  charged  for  each  reserved  seat 
ticket  and  35  cents  for  each  admission  ticket.  The  ticket  seller  showed  by 
his  account  that  500  tickets  were  sold,  for  which  he  received  $236.  How 
many  people  bought  reserved  seat  tickets  ? 

Let  X  stand  for  the  number  of  people  who  bought  reserved  seat  tickets 
at  $0.75  each.  Then  500  —  x  will  stand  for  the  number  of  people  who 
bought  admission  tickets  at  $0.35  each. 

By  the  conditions  of  the  problem  we  have 

.75  a:  +  .35  (500  -  x)  =  236. 

Solving,  we  obtain  x  =  152^. 

The  result  x  =  152^  satisfies  the  conditional  equation  but  not  the  im- 
plied conditions  of  the  problem,  since  it  is  impossible  to  give  a  sensible 

21 


322  FIRST  COURSE   IN  ALGEBRA 

interpretation  to  a  fractional  number  as  representing  tlie  number  of  people. 

It  appears,  then,  that  the  conditions  of  the  problem  as  stated  are  in- 
consistent when  applie<l  to  people. 

If  the  amount  received  had  been  given  as  $235  instead  of  $236  the 
number  of  people  would  have  been  found  to  be  ec[ual  to  150;  that  is,  we 
would  have  obtained  an  answer  which  could  have  been  given  a  sensible 
interpretation. 

21.  In  connection  with  the  interpretation  of  solutions,  the  follow- 
ing problem  has  become  classical : 

The  Problem  of  the  Couriers.  Two  couriers,  A  and  B,  are 
travelling  in  the  same  direction  along  the  same  road  at  the  uniform 
rates  of  m  miles  an  hour  and  n  miles  an  hour  respectively.  At  a 
specified  time,  say  at  noon,  B  is  c?  miles  in  advance  of  A.  Will 
they  ever  be  together,  and  if  so,  when  1 

Let  X  represent  the  number  of  hours  after  12  o'clock  when  they  will  be 
together. 

During  that  time  A  will  travel  mx  miles,  and  B  will  travel  tix  miles. 

Since  at  noon  B  is  d  miles  in  advance  of  A,  we  may  form  the  conditional 
equation  mx  =  7iar  4-  f?  (1),  of  which  the  solution  is  found  to  be 

x=^—    (2) 

m  —  n 

We  will  now  examine  this  expressed  value  of  x  and  determine  what 
restrictions,  if  any,  must  be  placed  upon  the  numbers  represented  by  d,  m, 
and  n  in  order  that  the  value  of  x  shall  be  consistent  with  the  stated 
conditions  of  the  problem. 

1.  If  A  is  to  overtake  B  after  12  o'clock,  the  value  of  x  must  be  jyositive. 
For  this  it  is  sufficient  that  fZ,  m,  and  n  all  represent  positive  numbers,  and 
that  m  >  n.  The  assumption  that  m  >  n  implies  that  A  is  travelling 
faster  than  B  and  accordingly  will  overtake  him. 

2.  If  m,  n,.and  d  were  all  positive  and  m  <  n,  the  value  of  x  would  be 
negative^  and  accordingly  we  should  interpret  the  negative  quality  of  x 
as  indicating  that  A  and  B  had  been  together  d/(n  —  m)  hours  before 
noon. 

3.  If  m  =  n,  then  m  —  n  =  0,  and  since  we  cannot  divide  any  number 
by  zero  it  appears  that  when  w  =  n,  we  cannot  obtain  the  solution  (2)  Irom 
the  given  equation  (1). 

If,  however,  m  instead  of  being  equal  to  n  differs  from  n  l)y  a  very  small 
amoimt,  the  difference  m  —  n  will  be  different  from  zero,  and  accordingly 


PROBLEMS  IN  PHYSICS  328 

the  smaller  the  value  of  m  —  n,  the  larger  for  a  given  value  of  d  will  the 
value  of  the  fraction  d/  (m  —  n)  become. 

This  is  commonly  expressed  by  saying  that  as  m  becomes  nearly  equal 
to  n,  X  becomes  infinite. 

This  slioukl  be  interpreted  as  another  way  of  saying  that  as  A's  rate 
becomes  more  and  more  nearly  equal  to  B's,  the  time  required  for  A  to  over- 
take B  will  become  correspondingly  greater  and  greater.  Finally  if  A's 
rate  is  equal  to  B's,  A  will  never  overtake  B. 

Accordingly,  an  infinite  solution  may  he  interpreted  as  meaning  that  it  is 
impossible  to  find  the  solution  under  the  assumed  conditions. 

4.  If  we  assume  "that  m  =  n  and  also  that  d  =  0,  the  equation  (1)  is 
satisfied  by  any  value  which  we  may  assign  to  x  and  the  general  solution 
assumes  the  indeterminate  form  x  =  0/0. 

This  may  be  interpreted  as  meaning  that,  since  the  distance  d  between 
A  and  B  is  zero,  and  their  rates  of  travelling,  m  and  n,  are  equal,  that  if  at 
any  time  they  are  together  they  will  always  be  together. 

22.  The  student  will  commonly  find  no  difficulty  in  .giving  sen- 
sible interpretations  to  the  solutions  of  particular  problems  whether 
these  solutions  be  positive  or  negative,  fractional,  zero,  indeterminate, 
or  infinite. 

Whenever  it  is  impossible  to  give  a  sensible  interpretation  to  any 
particular  solution  which  may  be  obtained,  it  will  be  a  good  exercise 
for  the  student  to  examine  the  data  of  a  given  problem  and  if  pos- 
sible to  ascertain  the  cause  of  the  inconsistency. 

Problems  in  Physics 

23.  The  Horizontal  Lever.  A  straight  horizontal  lever  at 
rest,  supported  at  some  point  F  called  the  fulcrum,  and  acted  upon 
at  distances  of  a  units  and  b  units  from  F  by  two  parallel  vertical 
forces  having  the  same  directions,  which  are  represented  numerically 
by  A  and  B,  will  remain  at  rest  provided  that  the  numbers  repre- 
sented by  a,  b,  A,  and  B  satisfy  the  conditional  equation  Aa  =  Bb. 

24.  The  product  obtained  by  multiplying  the  number  which 
represents  the  force  A  by  the  number  which  represents  the  distance 
a  of  the  point  of  application  of  the  force  from  the  point  of  support 
F,  is  called  the  moment  of  the  force  A  with  respect  to  the  point  of 
support  F. 

25.  Since  the  forces  A  and  B  tend  to  produce  rotation  of  the 
horizontal  bar  in  opposite  directions  about  the  fulcrum  i^'as  a  point 


324  FIRST  COURSE  IN  ALGEBRA 

of  support,  it  may  be  seen  that  the  condition  of  equilibrium  is  that 
the  moment  Aa  shall  be  equal  to  the  moment  Bb. 

See  Fig.  2,  in  which  vertical  forces  A  and  B  have  the  same 
direction  and  act  upon  the  horizontal  lever  at  points  situated  at 
distances  a  and  b  on  opposite  sides  of  the  fulcrum  K  In  Fig.  3 
the  forces  A  and  B  have  opposite  directions  and  act  on  the  hori- 
zontal lever  at  points  which  are  situated  on  the  same  side  of  the 
fulcrum  F. 

b 

. ^ .B 

aba  ^ 

I      i  i     5     I 

A  B  A 

Fig.  2.  Fig.  3. 

26.  A  horizontal  bar  in  equilibrium  will  remain  at  rest  if  vertical 
forces,  represented  by  ^,  By  (7,  D,  E,  etc.,  acting  at  distances  a,  6, 
c,  df  Sf  etc.,  from  the  fulcrum  F,  satisfy  a  conditional  equation  such 
as  Aa  +  Cc  =  Bb-\-  Dd  +  Ee. 

27.  It  is  a  principle  that  a  mass  may  be  treated  in  calculations 
as  if  it  were  concentrated  at  a  certain  point  called  the  center  of 
gravity  of  the  mass. 

Exercise  XVI.     8 

Solve  each  of  the  following  problems  : 

1.  How  heavy  a  stone  can  a  man,  by  exerting  a  force  of  160  pounds, 
lift  with  a  crowbar  6  feet  in  length,  if  the  fulcrum  be  one  foot  from  the 
stone  (neglecting  the  weight  of  the  crowbar)  ? 

2.  A  wheelbarrow  is  loaded  with  50  bricks,  each  weighing  6  pounds. 
What  lifting  force  must  be  applied  at  the  handles  to  raise  the  load  (neglect- 
ing the  weight  of  the  wheelbarrow),  provided  that  the  center  of  gravity 
of  the  load  is  2  feet  from  the  center  of  the  wheel  and  the  hands  are  placed 
at  a  distance  of  4  feet  from  the  center  of  the  wheel  ? 

3.  A  beam  20  feet  in  length  and  weighing  50  pounds  is  supported  at  a 
point  4  feet  from  one  end.  What  force  must  be  applied  at  the  end  farthest 
from  the  point  of  support  to  keep  the  beam  in  equilibrium  1  What  force 
must  be  applied  at  the  end  nearer  the  point  of  support  ? 

4.  A  board  15  feet  in  length  and  weighing  21  pounds  is  supported  at 
a  point  2  feet  from  the  center.  If  the  board  is  kept  in  equilibrium  by  a 
Btone  placed  on  it  at  a  point  3  feet  from  the  fulcrum,  find  the  weight  of  the 
stone. 


PROBLEMS  IN   PHYSICS  325 

5.  A  horizontal  bar  18  inches  in  length  is  in  equilibrium  when  forces 
of  4  pounds  and  2  pounds  respectively  are  acting  downward  at  its  ends. 
Find  the  position  of  the  point  of  support. 

6.  A  basket  weighing  100  pounds  is  suspended  at  a  point  two  feet  from 
the  end  of  a  stick  which  is  8  feet  in  length  and  which  weighs  three  pounds. 
If  the  stick  is  being  carried  by  two  boys,  one  at  each  end,  how  many 
pounds  does  each  boy  lift  ? 

7.  Two  boys,  one  at  each  end  of  a  stick  12  feet  in  length  which  weighs 
5  pounds,  raise  a  certain  weight  which  is  suspended  from  the  stick.  How 
heavy  is  the  weight  and  at  what  point  does  it  hang,  if  one  boy  lifts  35 
pounds  and  the  other  lifts  30  pounds  ? 

Since  the  boys  lift  35  and  30  pounds  respectively,  and  the  weight  of  the 
Gtick  is  5  pounds,  it  follows  that  the  weight  carried  must  be  60  pounds. 

If  the  stick  be  assumed  to  be  uniform,  it  may  be  seen  that  one  boy  will 
carry  32^  pounds  and  the  other  Ijoy  27^  pounds  of  the  weight. 

We  will  represent  by  x  the  number  of  feet  from  the  center  of  gravity  of 
the  stick  to  the  point  at  which  the  weight  is  suspended  on  the  side  of  the 
center  of  gravity  nearer  the  boy  exerting  the  greater  force. 

It  may  be  seen  that,  with  respect  to  the  center  of  gravity,  regarded  as  a 
fixed  point,  the  weight  of  60  pounds  which  is  carried  and  the  force  of 
27^  pounds  exerted  at  one  end  of  the  stick  both  tend  to  produce  rotation 
of  the  stick  about  its  center  of  gravity  in  one  direction,  while  the  force  of 
32^  pounds  exerted  at  the  other  end  of  the  stick  tends  to  produce  rotation 
of  the  stick  about  its  center  of  gravity  in  the  opposite  direction. 

Since  the  stick  is  in  equilibrium,  the  sums  of  the  moments  of  these  forces 
must  be  equal. 

Hence  we  have  the  following  conditional  equation: 
60a:  +  (27^)  x  6  =  (32^)  x  6. 

Solving,  we  obtain  x  =  l^. 

Hence,  the  weight  is  suspended  from  the  stick  at  a  point  which  is  6 
inches  from  the  center  of  gravity  on  the  side  nearer  the  boy  exerting  the 
greater  force. 

This  value  will  be  found  to  satisfy  the  conditions  of  the  given  problem. 

8.  A  safety  valve  having  an  area  of  4  square  inches  is  held  down  by  a 
lever  which  is  hinged  at  one  end. 

The  lever  is  10  inches  long  and  the  point  of  application  of  the  valve  is 
2  inches  from  the  hinged  end  of  the  lever.  If  a  weight  of  12  pounds  is 
placed  on  the  free  end,  find  the  pressure  per  square  inch  on  the  valve  which 
will  lift  the  safety  valve,  disregarding  the  weight  of  the  lever. 

9.  A  dog-cart  carrying  a  load  of  576  pounds  is  found,  when  on  a  level 
road,  to  exert  a  pressure  of  only  8  pounds  on  the  horse's  back.     If  the  dis- 


1 


326  FIRST  COURSE  IN  ALGEBRA 

tance  from  the  point  of  support  on  the  horse's  back  to  the  axle  be  6  feet, 
find  the  distance  of  the  center  of  gravity  of  the  load  from  the  axle. 

10.  A  board  of  uniform  thickness,  weighing  30  pounds,  is  balanced  when 
supported  at  a  point  4  feet  from  one  end  and  when  a  weight  of  70  pounds 
is  placed  one  foot  from  this  end.     Find  the  length  of  the  board. 

11.  A  beam  of  uniform  thickness,  20  feet  in  length,  is  supported  at  a 
point  8  feet  from  one  end.  If  the  beam  is  balanced  when  a  weight  of  80 
pounds  is  placed  on  the  end  nearer  the  fulcrum  and  a  weight  of  30  pounds 
is  placed  on  the  end  farther  from  the  fulcrum,  what  is  the  weight  of  the 
beam? 

Exercise  XVI.    9    Review 
Simplify  each  of  the  following  : 

6.  Find  the  remainder  when  6 a*  —  1  x^  +  5x^  —  2x  +  3  is 
divided  by  a  —  5. 

7.  Factor  10  (a^  +  1)  —  29  a. 

8.  Factor  x^  +  2  xy  -^  if  +  x  +  y. 

9.  Factor  {a  —  bf  -  8. 

10.  Find  the  L.  C.  M.  of  «*  +  «'  +  1  and  a^  -  a" -[-  1. 
Simplify  each  of  the  following  : 

12.  (a3-,i,)-^(a^  +  ia  +  ^.).    14.  |^-|±|. 
15.  L^ ^_,_. ±. 


16.  Show  that  (ar^  +  3a+ 2)(a;'+7a;+12)  =  (a!2+4a;+3)(ar'+6a;4-8). 

T^.    ,,,        1       p^  —  6+1      V  m  +  I       ,,      mn-\-m 

1 7.  Find  the  value  of  — --, » when  a  = -—  and  o  = — -  • 

a  +  6  —  1  mn  +  1  mn  +  1 

18.  Divide  the  product  of  a-\-b  —  c,  b  +  c  —  a  and  c  +  a  —  bhy 
a^-b''-c^-2bc. 


SIMULTANEOUS  EQUATIONS  827 


CHAPTER  XVII 

SIMULTANEOUS  LINEAR   EQUATIONS 

General  Principles  of  Equivalence 

1.  Two  or  more  conditional  equations  are  said  to  be  simul- 
taneous with  reference  to  two  or  more  unknowns  appearing  in 
them  when  each  unknown  letter  is  assumed  to  represent  the  same 
number  wherever  it  appears  in  all  of  the  equations. 

2.  A  set  or  group  of  simultaneous  equations  is  called  a  system 
of  simultaneous  equations. 

E.  g.  The  equations  3  x  +  2  y  =  14  (1)  and  a:  +  5  y  =  9  (2)  are  simul- 
taneous on  condition  that  x  represents  the  same  number  in  (1)  as  in  (2), 
and  that  y  lias  the  same  value  in  one  equation  as  it  has  in  the  other. 

3.  A  solution  of  a  conditional  equation  containing  two  or 
more  unknowns  is  any  set  of  values  which,  when  substituted  for  the 
unknowns,  reduces  the  conditional  equation  to  an  identity. 

E.  g.  The  sets  of  two  values,  a:  =  2,  2/  =  4;  a:  =  0,  i/  =  7;a:  =  6,  i/  =  ~2, 
etc.  ;  are  solutions  of  the  conditional  equation  3a:  +  2i/=  14  containing  two 
unknow^ns. 

4.  A  solution  of  a  system  of  simultaneous  conditional  equa- 
tions is  any  set  of  values  of  the  unknowns  which  satisfies  all  of  the 
equations  of  the  system. 

E.  g.  The  single  set  of  two  values  rr  =  4,  ?/  =  1  is  the  single  solution  of 
the  system  of  two  simultaneous  conditional  equations  3x-{-2y=  14  (1), 
and  x  +  by  =  i)  (2). 

6.  The  word  "  solution  "  may  be  used  to  denote  either  the  pro- 
cess of  solving  an  equation  or  system  of  equations,  or  the  value  or 
values  obtained  by  the  process. 

6.  If  the  number  of  solutions,  —  that  is,  the  number  of  different 
sets  of  values  which  satisfy  all  of  the  equations  of  a  given  system 
of  simultaneous  conditional  equations,  —  is  limited  or  finite,  the 
system  is  said  to  be  determinate. 


328 


FIRST  COURSE   IN  ALGEBRA 


If,  however,  the  number  of  different  sets  of  values  which  satisfy- 
all  of  the  equations  of  a  system  be  unlimited  or  infinite,  the  system 
is  said  to  be  indeterminate. 

7.  Two  conditions  restricting  the  values  of  two  or  more  unknown 
numbers  are  said  to  be  consistent  if  both  conditions  can  be  satis- 
fied by  the  same  values  of  the  unknowns. 

In  the  contrary  case,  the  conditions  are  said  to  be  inconsistent. 

E.  g.  If  we  are  required  to  find  two  numbei-s  whose  suiii  is  10  and 
difference  8,  tlie  conditions  restricting  the  vahies  of  the  unknown  numbers 
ai-e  consistent,  since  we  can  find  two  numbers,  9  and  1,  which  satisfy  them. 

Two  conditions  requiring  that  the  sum  of  two  unknown  numbers  shall 
be  10  and  also  8  are  inconsistent,  since  it  is  impossible  to  find  two  such 
numbers. 

8.  The  i^rrapli  of  a  conditional  equation  of  the  first  degree 
containing  two  unknowns  is  a  straight  line. 

This  may  be  shown  directly  by  applying  certain  simple  properties  of 
plane  triangles. 

(The  following  proof  is  offered  for  such  students  as  are  acquainted  with  a  few  of  the  simple 
principles  of  geometry,  and  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  A  and  A'  represent  any  two  points  on 
the  graph  of  a  given  equation  y  =  ax,  located 
by  means  of  the  coordinates  (x,  y)  and  (x',  y'), 
so  taken  that  corresponding  values  of  x  and  y 
satisfy  the  given  equation.     (See  Fig.  1.) 

Draw  straight  lines  from  the  origin  0  to  A 
and  A'. 

Since  the  values  a;,  y,  and  x',  y\  are  assumed 
to  satisfy  the  equation  y  =  oa:,  we  have  y  =  ax 
and  y'  z=:  aj/. 

V  1  y' 

-  =  a.  and  ^  =  a. 

X        '  x' 


T 

^ 

A 

y 

f 

0 

X        B 

px 

r 

Fig.  1. 


Hence 


Therefore 


The  ordinates  y  and  y'  are  taken  parallel  to  the  axis  of  F,  and  accord- 
ingly the  triangles  DBA  and  OB' A'  are  similar,  since  they  have  an  angle 
of  one  equal  to  an  angle  of  the  other,  and  the  included  sides  proportional. 

The  corre.sponding  angles  AOB  and  A' OB'  are  consequently  equal,  and 
the  lines  OA  and  OA'  coinpide. 


SIMULTANEOUS  EQUATIONS 


329 


It  follows  that  either  of  the  points  A  or  A'  lies  on  the  straight  line 
drawn  from  the  origin  to  the  other  point,  and  since  A  and  A'  represent 
any  two  points  on  the  graph,  they  represent  all  points  on  the  graph,  which 
must  accordingly  be  a  straight  line  passing  through  the  origin. 

It  may  be  observed  that  the  inclination  of  the  line  with  the  a:-axis 
depends  wholly  upon  the  value  of  a,  since  for  a  given  value  of  x  the  length 
of  y  is  equal  to  the  product  ax.  The  line  will  slope  upward  or  downward 
toward  the  right  according  as  a  is  positive  or  negative. 

The  graph  of  the  equation  y  =  ax  +  b  may  be  obtained  by  adding  b  to 
each  of  the  ordinates  calculated  for  the  graph  of  the  equation  y  =  ax.  The 
x-coordinates  will  be  the  same  for  the  graphs 
of  both  of  the  equations,  but  the  y-ordinates  of 
the  graph  of  the  equation  y  =  ax  +  b  will  be 
greater  by  b  than  those  of  the  graph  of  the 
equation  y  =  ax. 

It  may  be  seen  that  the  figure  AA'A"A'"  is 
a  parallelogram  by  construction,  and  accord- 
ingly the  straight  line  A'" A"  is  parallel  to  the 
straight  line  AA'.     (See  Fig.  2.) 

Accordingly,  the  graph  of  the  equation 
y  =  ax  +  b  is  a  straight  line  which  is  parallel 
to  the  graph  of  the  equation  y  =  ax.     (Conqjare  with  Chapter  IX.  §  37.) 

9.  Since  the  graph  of  every  equation  of  the  first  degree  contain- 
ing two  unknowns,  such  as  x  and  y,  is  a  straight  line,  an  equation 
of  the  first  degree  with  reference  to  the  unknowns  appearing  in  it 
is  commonly  called  a  simple  or  linear  equation,  that  is,  the 
equation  of  a  line, 

10.  To  obtain  gi'aphically  the  solution  of  a  system  of  two 
linear  equations  containing  two  unknowns,  we  plot  the  graphs 
representing  the  equations  and,  locating  their  intersection,  if  there  he 
one,  measure  the  x-coordinate  and  y-coordinate  corresponding  to  this 
point  and  estimate  the  corresponding  numerical  values,  attaching  the 
proper  quality  signs  determined  by  the  quadrant  in  which  the  point 
lies. 

11.  Since  any  two  straight  lines  lying  in  the  same  plane,  which 
are  not  coincident,  must  either  intersect  or  be  parallel,  it  follows 
that  pairs  of  equations  of  the  first  degree  containing  two  unknowns 
may  be  separated  into  three  classes  :  one  class  consisting  of  such 
pairs  of  equations  as  are  represented  graphically  by  intersecting 


830  FIRST  COURSE  IN  ALGEBRA 

straight  lines  {Independent  equations) ;  a  second  class  consisting  of 
those  pairs  of  equations  which  are  represented  by  lines  which  do  not 
meet,  that  is,  which  are  parallel  (inmnsistent  equations);  and  a  third 
class  consisting  of  those  pairs  of  equations  which  may  be  reduced  to 
exactly  the  same  form,  that  is,  which  represent  coincident  lines 
(equivalent  equations). 

12.  Two  or  more  conditional  equations  which  express  different 
consistent  conditions  restricting  the  values  of  the  same  unknowns 
are  called  independent  equations. 

Of  two  equations  which  are  independent,  neither  can  be  trans- 
formed into  the  other. 

E.  g.  The  two  conditional  equations  2x  -\-y  =  12  and  3x  +  7  y  =  2d  are 
independent,  since  neither  am  by  any  transformation  be  made  to  take  the 
form  of  the  other. 

13.  The  point  of  intersection  of  the  graphs  of  two  linear  equa- 
tions containing  two  unknowns  may  be  located  by  means  of  the  two 
coordinates  which  are  equal  tx)  the  values  found  by  solving  alge- 
braically the  two  equations  of  which  they  are  the  graphs. 

E.  g.  Consider  the  two  independent  linear 
equations  x  -\-2y  =  S  and  3x  —  y  =  3. 

Any  set  of  values  satisfying  either  equa- 
tion may  be  taken  as  coordinates  locating 
a  definite  point  on  the  graph. 

Hence  the  values  x  =  2,  y  =  3,  which 
are  the  common  solution  of  the  two  given 
equations,  are  equal  to  the  coordinates 
X  =  2  and  i/  =  3  of  the  point  common  to 
the  two  lines  in  Fig.  3. 

14.  Two  or  more  conditional  equations  which  express  consistent 
conditions  existing  between  the  unknowns  appearing  in  them  are 
called  consistent  equations. 

E.  g.  The  two  conditional  equations  x -\- y  =  12  and  x  —  y  =  6  are  con- 
sistent since  both  are  satisfied  by  the  values  x  =z  9  and  y  =  3. 

16.  Consider  the  conditional  equations  x  +  7/  =  1,  x  —  y  =  1^ 
3 ar  +  2  ?/  =  18,  and  ic  —  4?/  =  —  8. 

Each  of  these  equations  is  satisfied  by  the  values  ic  =  4  and  y  =  S, 


;' 


SIMULTANEOUS   EQUATIONS 


331 

Any  two  of 


^ 


j:+y=7 


Fjg.  4. 


and  the  equations  are  all  independent  and  consistent. 
them  will  serve  to  determine  the  values  of  x  and  y. 

Referring  to  the  accompanying  figure,  in  which  portions  of  the 
graphs  of  these  equations  are  plotted,  it  appears  that  these  equa- 
tions represent  separate  straight  lines, 
all  passing  through  a  common  point  A 
whose  coordinates,  ic  =  4  and  2/  =  3,  are 
equal  to  the  common  solutions  of  the 
equations. 

From  the  illustration  it  appears  that, 
since  any  particular  point  such  as  A  is 
located  definitely  by  means  of  any  two 
straight  lines  passing  through  it,  and 

not  more  than  two  lines  are  necessary  to  locate  the  point,  so  two 
indejjendsnt  conditional  equations  of  the  first  degree  containing  two 
unknmvns  determine  the  values  of  two  unkrwwn  numbers^  and  more 
than  two  equatiims  are  unnecessary. 

16.  Two  conditional  equations  which  express  inconsistent  con- 
ditions restricting  the  values  of  the  unknowns  appearing  in  them 
are  called  iuconsisteut  equations. 

E.  g.  The  conditional  equations  oj  -f-  7/  =  7  and  a;  -f-  7/  =  5  are 
inconsiatent. 

17.  Two  or  more  inconsistent  equations  can  have  no  solution  in 

common. 

E.  g.    Consider  the  two  conditional  ec^uations 

3x4-21/  =  6  and  3x-i-2y=l2. 

Since  it  is  impossible  that  3x  -\-2ij  should 
equal  6  and  also  12  at  the  same  time,  these  must 
be  classed  as  inconsistent  equations. 

On  attempting  to  solve  the  equations  as  sim- 
ultaneous equations  we  shall  find  that  they  have 
no  common  solution. 

If  their  graphs  are  plotted  with  reference  to 
the  same  axis  of  reference  we  shall  find  that 
they  appear  to  be  parallel  straight  lines.  (See 
Fig.  5.) 

18.  Since  the  point  of  intersection,  if  there  be  one,  of  the  graphs 
of  two  linear  equations  containing  two   unknowns  is  located  by 


3x+2y=6 


3x+2y=12 


Fig.  6. 


332 


FIRST  COURSE   IN  ALGEBRA 


means  of  coiirdinates  equal  to  the  values  which  form  the  common 
solution  of  the  given  equations,  it  follows  that  if  the  equations  have 
no  common  solution  their  graphs,  which  are  straight  lines,  can  have 
no  point  in  common,  and  accordingly  must  be  parallel  straight  lines. 
19.  Two  conditional  equations  containing  two  or  more  unknowns 
are  said  to  be  equivalent  when  every  solution  of  either  equation  is 
at  the  same  time  a  solution  of  the  other  equation ;  that  is,  when  any 
set  of  values  satisfying  either  equation  satisfies  the  other  equation 
also.     (See  also  Chap.  X.  §  22.) 

E.  g.  Since  two  conditional  equations  are  equiv- 
alent when  each  equation  is  satisfied  by  all  of  the 
solutions  of  the  other,  it  foHows  that  the  graphs  of 
two  equivalent  equations  such  as  5  a;  +  3  y  =  15 
and  10;c  +  62/  =  30  must  contain  the  same  ])oint8, 
and  neither  graph  can  contain  any  point  which  the 
other  does  not. 

Hence  the  graphs  must  be  coincident  lines.  (See 
Fig.  6.) 

20.  It  may  be  observed  that  independent 
conditional  equations  express  different  consist- 
ent relations  between   the   unknowns,   while 

equivalent    equations    express    the    same    relations   between    the 

unknowns. 

E.  g.  Any  one  of  the  following  equations  is  equivalent  to  any  other, 
since  of  any  pair  of  equations  either  equation  may  be  transformed  into  the 
other : 

3x  +  2  2/ =  7,  6x  +  41/ =  14,  2x  +  5^/ =  33/ -  X  +  7,  ^ -}- -^^  =  1. 


Fig.  6. 


Exercise  XVII.     1 

Of  the  following  equations  select  those  which  are  equivalent  to 
the  equation  2  a;  +  3  ^  =  10  : 

1.  4a;  +  63/ =  20.  3.  3x+  S^j=  10  —  a;. 

2.  6a;+  12?/ =  30.  4.  x  + -^  =  5. 


5.  2{x  +  ^j+  1)  =  12-^. 


SIMULTANEOUS  EQUATIONS  e333 

Of  the  following  equations  select  those  which  are  equivalent  to 
the  equation  4  cc  —  2 1/  =  3  ; 

6.   12a; +  8?/ =  9.  8.  2a;-^  =  |. 


7.  8a; -4^  =  6.  ^-  |  ~  4  = 


10.  20a;—  10?^=  16. 
Among  the  following  sets  of  equations, 

(i.)  which  have  equations  which  are  independent  and  consistent  1 
(ii.)  equivalent?  (iii.)  inconsistent? 

11.  2a!  +  3 ^/  =  10,  n.Sx  —  22/  =  0, 
4a;  +  62^  =  20.  2a;-  37/=0. 

12.  5x—  73^  =  9,  18.  4a;  +  5?/  =  6, 
2a;+     ?/  =  7.  5a; +63^  =  7. 

13.  3a; +  8^=  12,  19.   10a;+  8  7/  =  3, 
6a;+16y  =  22.  5a;  -  4?/  =  6. 

14.  a;  +  ?/  -  2  =  0,  20.  2  a;  +     ^  =  3, 

X  —  1/ =  I.  a; +3?/ =  2. 

15.  3a;  — y=12,  21.  2a;  —  3  =  4?/, 

?/         ,  2v/  — 3  =  4a;. 

a;  —  -  =    4. 
3        *• 

16.  12a;- 9y  =  18,  22.5+     x  =  6^, 

8x  —  6i/=U.  1/  +  5x=6. 

21.  It  can  be  shown  that  the  necessary  and  sufficient  condition 
that  a  system  of  simultaneous  linear  equations  shall  have  a  definite 
number  of  solutions  is  that  there  shall  be  the  same  number  of  in- 
dependent and  consistent  equations  as  there  are  unknowns  whose 
values  are  to  be  found. 

22.  Two  systems  of  equations  are  equivalent  when  every  solu- 
tion of  either  system  is  also  a  solution  of  the  other. 

E.  g.     The  equations  in  groups  I.  and  II.  below  form  equivalent  systems, 
for  the  equations  in  either  <,'roup  are  satisfied  by  the  solution  x  =  I  and  ^ 
y  =  3  ;  we  shall  show  later  that  they  are  satisfied  by  no  other  solution. 

2.  +  5,  =  17,  I  j_  f  +  ^''  =  "'  I  System  II. 


334  FIRST  COURSE  IN  ALGEBRA 

Elimination 

23.  Any  process  by  means  of  which  the  members  of  two  or  more 
equations  may  be  combined  to  produce  a  derived  equation  in  which 
fewer  unknowns  appear  than  in  the  equations  whose  members  have 
been  combined  to  produce  it,  is  called  a  process  of  elimination. 

24.  Any  unknown  which  is  found  in  two  or  more  given  equa- 
tions but  which  does  not  appear  in  an  equation  derived  from  them 
is  said  to  have  been  eliminated. 

25.  The  general  principles  governing  the  derivation  of  equivalent 
equations  containing  one  unknown,  proved  in  Chapter  X.,  hold  true 
also  for  equations  containing  any  number  of  unknowns. 

26.  The  processes  of  elimination  employed  in  the  solution  of  a 
system  of  simultaneous  linear  equations  may  be  made  to  depend 
upon  the  following 

General  Principles 

Principle  (i.)  If^for  any  equation  of  a  system  of  simultaneotis 
eqtuitions  an  equivalent  eqtiation  be  substituted,  the  system  composed 
of  this  derived  equation,  taken  together  with  the  remaining  original 
equations^  will  be  equivalent  to  the  original  system  of  equations. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  a  system  of  simultaneous  equations  containing  two  unknowns,  say  x 

and  ?/,  be  re]3resented  by 

A  —  n  ^ 

Given  System. 


^  =  (7,) 

B  =  D.  i 


Let  the  equation  />'  =  D'  be  derived  from  the  equation  B  =  D,  and  let 
^  =  D'  be  equivalent  to  B  =  D. 

Then  the  system  composed  of  the  other  given  equation,  A  =  C,  and  the 
derived  equation,  B'  =  U,  will  be  equivalent  to  the  original  System  I.,  for 
by  the  definition  of  e(iuivalent  equations,  the  equivalent  equations  B  =  D 
aud  B'  =  ly  have  the  same  solutions. 

A  =  C,   "^  Equivalent 

V  IL         Derived 
B'  =  D'.  )  System. 

Hence,  any  set  of  values  whicli  satisfies  either  of  the  equations  B  =  D 
or  B'  =^  D'  and  also  the  e<£uation  A  =  C  must  satisfy  both  the  given  and 


SIMULTANEOUS   EQUATIONS  B35 

the   derived  systems  ;    that   is,  the  systems  of  equations    I.    and    II.    are 
equivalent. 

The  reasoning  may  be  extended  to  include  systems  of  three  or  more 
equations  containing  three  or  more  unknowns. 

27.  In  elementary  algebra,  methods  of  elimination  by  substitu- 
tion, by  comparison,  and  by  addition  or  subtraction  are  commonly 
employed. 

I.    Elimination  by  Substitution 

28.  The  method  of  elimination  by  substitution  may  be  made  to 
depend  upon  the  following 

Principle  (ii.):  If  in  any  equation  belonging  to  a  system  of 
simultaneous  equations  the  value  of  one  of  the  unknowns  be  expressed 
in  terms  of  the  remaining  unknown  numbers  and  known  numbers 
appearing  in  the  same  equation^  and  this  expressed  value  thus  ob- 
tained be  substituted  for  the  same  unknown  wherever  it  appears  in 
the  remaining  equation  or  equations  of  the  system^  then  the  derived 
system  will  be  equivalent  to  the  given  system. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

We  will  represent  the  linear  equations  composing  a  given  system  of  two 
simultaneous  equations,  in  which  two  unknowns,  x  and  i/,  appear,  by 

B  =  D.     (2)  j  ^^^^^  System. 

Representing  by  E  the  expressed  value  of  one  of  the  unknowns,  say  y, 
in  terms  of  the  remaining  unknown  ar,  and  of  the  known  numbers  appearing 
in  one  of  the  equations,  —  say  equation  (1),  —  we  may  derive  the  equivalent 
equation 

y=E.     (3) 

Substituting  this  expressed  value,  E,  for  y  wherever  y  is  found  in  the 
remaining  equation,  —  say  equation  (2)  of  System  I., —we  may  represent 
the  derived  equation  by 

B'  =  D'.     (4) 

We  are  to  show  that  System  II.,  composed  of  equations  (3)  and  (4),  is 
equivalent  to  the  given  system  of  equations  (1)  and  (2),  that  is,  to  System  I. 
y  =  E,      (3)  )  Equivalent 

>-  II.         Derived 
B'  =  D'.     (4)  )  System. 


336  FIRST  COURSE   IN  ALGEBRA 


Since  y  =  E  18  equivalent  to  equation  (1),  the  system  composed  of  this 
equation  and  the  remaining  original  equation  (2),  that  is,  System  III.,  must 
be  equivalent  to  the  original  System  I. 

y  =  Et      (3)  {  Equivalent 

)  III.         Derived 
B  =  D.       (2)  (  System. 

Any  solution  of  System  III.  satisfies  equations  (2)  and  (3),  that  is, 
makes  each  of  them  an  identity.     Hence  we  have 

y  =  E,  and  B  =  D. 

Any  solution  therefore  which  satisfies  (3)  and  (2)  must  also  satisfy  (2) 
after  E  has  been  substituted  for  y,  that  is,  must  satisfy  the  equation 
^  =  1/.     (4) 

It  follows  that  any  solution  of  System  III.  is  also  a  solution  of 
System  II. 

Furthermore,  any  solution  of  System  II.  makes  y  =  Ej  and  also  B'  =  D'. 

Hence,  any  solution  of  System  II.  must  also  satisfy  B' =  D'  after  y 
has  been  substituted  for  E^  that  is,  must  satisfy  tlie  equation  B  =  D  (2). 

Also,  since  the  equation  y  =.  E  \8  equivalent  to  the  equation  A  =  C\  any 
solution  of  System  II.  must  satisfy  also  System  I. 

Hence  Systems  I.  and  II.  are  equivalent. 

The  reasoning  may  be  extended  to  include  a  system  of  three  or  more 
equations  containing  three  or  more  unknowns. 

Systems  of  Linear  Equations  containing  two  Unknowns 

29.  The  method  of  elimination  by  substitution  may  be  used  to 
advantage  whenever  one  of  the  given  equations  contains  a  single 
unknown. 

Ex.  1.     Solve  the  following  system  of  equations  : 

o"^"!    ".o'     921^-     Given  System. 
3a; +  51/ =  23.     (2)  j  ^ 

From  equation  (1)  we  obtain  directly 

£c  =    6.     (3) ^         E     •     1     t 

Substituting  this  value  for  x  in  equation  (2)  we  have    (  r)    '      1 

y=  1-   (4) -'  ' 

Hence  the  solution  of  System  II.,  and  consequently  of  System  I.  is 


a:  =  6,  I 
?/=l| 


SIMULTANEOUS   EQUATIONS 


337 


In  Fig.  7  portions  of  the  graphs  of  the  equations  a:  —  6  =  0  and 
3a:  +  5^  =  23  are  sliovvn,  and  the  intersection  of  these  graphs  is  a  point, 
of  which  the  numerical  values  of  the  coor-  ^ 
dinates  are  equal  to  the  values  of  x  and  ?/, 
found  by  solving  the  given  equations. 

The  student  may  establish  the  equiva- 
lence of  the  equations  in  Systems  I.  and  II. 

We  may  verify  the  solution  by  substi- 
tuting the  values  6  and  1  for  x  and  ij 
respectively,  in  the  original  equations  (1) 
and  (2). 

Substituting  in  (I), 

6-6  =  0 
0  =  0. 


Fig.  7. 

Substituting  in  (2), 
18  +  5  =  23 


23  =  23. 

Ex.  2.    Solve  the  system  of  simultaneous  equations 
2a;+     v=  15,       (1)  )  ^      ^. 
5x-2,=    6.       \i)V'     <^-- System. 

"We  are  to  find  a  value  for  x  and  also  one  for  y  which  wall  satisfy  both 
of  the  given  equations. 

Although  we  do  not  at  fii-st  know  the  values  of  x  and  ?/,  we  proceed  upon 
the  assunq)tion  that  each  letter  has  the  same  value  in  one  equation  that  it 
has  in  tlie  other. 

From  equation  (I)  we  may  obtain  the  expressed  value  of  y  in  terms  of 
X  and  the  numbers  entering  into  the  equation. 

That  is,  i/=15-2a:.  (3) 

Substituting  15  —  2  a:  for  y  in  equation  (2),  we  obtain  the  following  de- 
rived equation  in  which  y  does  not  appear  : 

5a:-2  (15-2x)  =  6.  (4) 

The  derived  System  II.,  composed  of  equations  (3)  and  (4),  is  equivalent 

to  the  original  system. 

,  ^  \  Equivalent 

5..-2(15-2x)  =  6.     (4)^  gy^^^^_ 

From  equation  (4)  we  obtain,  a:  =  4. 

Substituting  this  value  in  equation  (3),  we  obtain  ^  =  7. 

Hence,  the  solution  of  System  II.,  and  consequently  of  System  I.,  is 


;} 


338  FIRST  COURSE  IN  ALGEBRA 


Fig.  8. 


In  Fig.  8  portions  of  the  graphs  of  the  ecjuations 
2x  +  y  =  15  and  5  a:  —  2y  =  6  are  shown,  and  the 
intersection  of  these  graphs  is  a  point,  of  which  the 
numerical  vahies  of  the  coordinates  are  equal  to 
the  values  of  x  and  y,  found  by  solving  tlie  given 
equations. 

We  may  verify  the  solution  by  substituting  the 
values  4  and  7,  for  x  and  y  respectively,  in  the  origi- 
nal equations  (I)  and  (2). 

Substituting  in  (1),  Substituting  in  (2), 

2-4  +  7=15  5-4-2-7  =  6 

15  =15.  6  =  G. 


The  equivalence  of  the  equations  employed  in  the  process  of  solution 
may  be  established  as  follows : 

Equation  (3)  is  equivalent  to  equation  (I)  by  Chap.  X.  §  27  (i.).  The 
system  composed  of  equations  (4)  and  (3)  is  e([uivalent  to  the  system 
composed  of  equations  (2)  and  (1)  by  §  26. 

It  follows  that  the  solution  of  System  II.  nmst  be  the  solution  of 
System  I. 

30.  The  general  method  of  solution  may  be  stated  as 
follows  : 

Obtain  from  one  of  the  equations  the,  expressed  value  of  one  of  the 
unknowns  in  terms  of  the  otJier ;  substitute  this  expressed  value  for 
tJie  same  unknown  wherever  it  appears  in  the  remaining  equation 
of  the  system^  and  solve  the  7'esulti?ig  equation. 

The  value  of  the  remainiiig  unknown  may  be  found  by  substituting 
the  value  of  the  unknown  just  found,  either  in  one  of  the  original 
equations  or  i?i  the  exp^r^ssed  value  of  the  other  unknown. 

31.  An  objection  to  this  method  is  that,  unless  the  coefficient 
of  the  unknown  to  be  eliminated  is  unity  in  the  equation  from  which 
the  expressed  value  of  this  unknown  is  obtained,  it  may  happen  that 
fractions  are  introduced  into  the  derived  equation. 

32.  The  preceding  examples  illustrate  the  principle  that  the 
solution  of  a  system  of  simultaneous  equations  which  consists  of  as 
many  equations  as  there  are  unknown  numbers  is  finally  made  to 
depend  upon  the  solution  of  a  single  equation  containing  a  single 
unknown. 


SIMULTANEOUS  EQUATIONS  339 

Exercise  XVIL     2 

Solve  each  of  the  following  systems  of  equations  by  the  method 
of  substitution,  verifying  all  results  numerically  : 

1.      ic4-9?^==19,  11.   12ic— 17v/- 2  =  0, 

Hx—     i/=ld.  X—       7/  —  1  =  0. 

2.x+nij=    0,  12.  4a;  4- 2.y- 12=^0, 

x-\-       .i/=lO.  x+     1/—    4:  =  0. 

3.  3ic-2?/=    4,  13..iB  =  3?/  — 2, 

5x+     ;i/  =  lh  7/  =  '6x  +  2. 

4:.     X—    y  =    8f  14.  ic  =  y, 

5a;+  6y  =  51.  8a;  +  3y=  11. 

5.  2a;  =12,  15.      x—    y  =  0, 

3a;  +  51/ =  53.  2x+  Sij  =  5. 

6.  3a;—    y=    6,  16.  25a;—    Qi/=    3, 

x-\-  9?^  =  86.  5a;  +  21 1/ =  10. 

7.  2  a;  4- ^  =  21,  17.  5  a;  —    ?^  —    5  =  0, 
2^-|-a;=  12.  7a;  +  3?/  — 24  =  0. 

8.  3a;— 12?^  =  0,  18.        x  +  i/  =  2a, 

a;  —    2  v/  —  14  =  0.  (f*  —  ^)^  =  (a  +  b)i/. 

9.  4  a;  =  3  ?/,  19.  ca;  —  %  =  0, 

7  a;  =  5  7/  +  1.  bx  —  cy  =  a. 

10.     7  a;  =  4  ?^,  20.  aa;  +  .y  =  ^, 

10a;  =  3  ?/  +  19.  a;  +  c^/  =  t?. 

33.  The  method  of  elimination  by  substitution  is  sometimes  em- 
ployed in  a  special  form  called  the  method  of 

dimiuation  by  Comparison 

An  unknown  is  eliminated  by  the  method  of  comparison  by  ob- 
taining from  each  of  two  given  equations  the  expressed  value  of  this 
unknown,  and  then  constructing  an  equation  the  members  of  which 
are  the  two  expressed  values  thus  obtained. 


340 


FIRST  COURSE  IN  ALGEBRA 


Ex.  1.   Solve  the  system  of  equations 

3a: +  41/ =  40,     (1)  ) 
9x-5y=    1.     (2)j 


I.    Given  System. 


From  equation  (1)  we  obtain  the  expressed  value  of  x, 
40-4y 


x  = 


(3) 


From  equation  (2)  we  obtain  the  expre: 
""-       9 


ise«l  value  of  a?,  }-  II 


(^) 


Equivalent 
Derived 
System. 


Since  these  expressed  values  of  x  represent  the  same  number,  they  may 
be  used  as  members  of  an  equation  ;  or,  from  another  point  of  view,  we 
may  substitute  for  x  in  one  of  the  equations,  say  (4),  the  expressed  value  of 
X  from  the  other  equation,  say  (3),  and  obtain 


40-42/ 
3 


1+5?/ 
9 


Hence,  2/  =  7 

From  either  equation  (3)  or  etjuation  (4)  we  may 

obtain  x  by  substituting  7  for  y. 

1  +  5  •  7 
From  equation  (4),  x  = 


Hence, 
The  values 


x  =  4 


(5) 


E([uivalent 
III.      Derived 
System. 


y=7,f 


are  the  solution  of  the  given  equations  (1)  and  (2), 


and  by  Principles  (i.)  and  (ii.),  the  systems  of  equations  I.,  II.  and  III.  are 
equivalent. 

Hence,  the  solution  of  the  given  System  I. 
is  the  single  solution  of  System  III. 

In  Fig.  9,  portions  of  the  graphs  of  the  given 
equations  3 a;  +  4 ^  =  40  and  9x  —  5y  =1 
are  shown,  and  the  numerical  values  of  the 
coordinates  x  =  4  and  y  =7  of  the  point  of 
intersection  of  the  graphs  are  equal  to  the  nu- 
merical values  of  x  and  y  found  by  solving  the 
Fig.  9.  algebraic  equations. 

The  solution  may  be  verified  by  substitut- 
ing the  values  4  and  7  for  x  and  y  respectively  in  the  given  equations  (1) 
and  (2). 


SIMULTANEOUS  EQUATIONS  341 

Substituting  in  (1),  Substituting  in  (2), 

3. 4  +  4-7  =  40  9-4-5-7  =  l 

40  =  40.  1  =  1. 

34.  The  preceding  example  illustrates  the  following  rule  for  elimi- 
nation by  comparison  : 

From  each  of  two  given  eqitations  find  the  expressed  value  of  one 
of  the  unknowns  and  foi'm  an  equation  the  members  of  which  are 
these  expressed  values. 

Solve  the  equation  thus  obtained  for  the  single  unknown  appearing 
in  it.  The  value  of  the  remaining  unknown  may  be  found  by  substi- 
tuting the  value  of  the  first  unknown^  thus  obtained,  for  the  first 
unknown  wherever  it  appears,  either  in  one  of  the  original  equations 
or  in  the  expressed  value  of  the  remaining  unknown. 

Exercise  XVII.     3 

Solve  the  following  systems  of  equations,  eliminating  the  unknown 
numbers  by  the  method  of  comparison  : 


1. 

a;  =  31/ -4, 

7. 

y  =  2x+l, 

35  =  4^-7. 

y  =  Sx-5. 

2. 

5a;- 2?/=  11, 

8. 

Sx-4y-  19  =  0, 

2  «  —  3  y  =    0. 

7a;  +  27/ -50  =  0. 

3. 

3a;  +  8y=  19, 

9. 

5a;+6y=    7, 

lx—2y=    1. 

8a;  +  9?/ =10. 

4. 

8aj=6y, 

10. 

11a;-    9y=    7, 

10a;=:27y  — 4. 

9  a;-  10  y=  11. 

5. 

3a; +7^  =  42, 

11. 

bx  -{-  ay  =  2  ab, 

5a;+  6y  =  53. 

ax-\-by  =  a^  +  b\ 

6. 

2 35-  5?/  =  —  23, 

12. 

X  +  ay  +  a^  =  0, 

3a;-4y  =  —    3. 

x  +  by  +  b^  =  0. 

11.   Elimination  by  Addition  or  Subtraction 

35.  The  method  of  elimination  by  addition  or  subtraction  may  be 
made  to  depend  upon  the  following 

Principle  (iii.)  If ,  for  any  equation  of  a  system  of  simultaneous 
equations,  an  equation  be  substituted  which  is  derived  from  two  or 


342  FIRST  COURSE  IN  ALGEBRA 

more  of  the  given  equations  of  the  system  by  either  adding  or  sub- 
tracting the  corresponding  members  of  these  eqtiations,  the  resulting 
system  will  be  equivalent  to  the  one  given. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 
If  d  and  n  represent  any  finite  numbers,  d  being  diflferent  from  zero,  the 
System  I.  is  equivalent  to  the  derived  System  II. 

'J  dA  -\-nB  =  0,  (3) ")  Equivalent 

li;  ,.U^-    Given  System  C  II. 

~       '^"M  B  =  0.  (2)  )         Derived  System. 

Every  solution  of  System  I.,  that  is,  every  set  of  values  making  both  A 
and  B  zero,  must  make  both  dA  and  nB  zero,  and  accordingly  must  satisfy 
System  II. 

Furthermore,  any  set  of  values  M'hich  satisfies  System  II.,  making  B  and 
dA  +  7iB  zeroj  must  make  dA  zero;  and  since  d  is  different  from  zero,  the 
remaining  factor  A  of  the  term  dA  must  be  zero. 

Hence  A  and  B  must  both  become  zero  for  any  particular  set  of  values 
which  satisfies  System  II. 

Accordingly,  such  a  set  of  values  must  satisfy  System  I.,  that  is,  Systems  I. 
and  II.  must  be  equivalent. 

36.  The  elimination  of  unknowns  by  the  method  of  addition  or 
subtraction  will,  in  the  majority  of  cases,  be  found  to  be  more  con- 
venient than  the  method  of  elimination  by  substitution.  This  is 
because  fractions  are  not  introduced  into  the  derived  equations  dur- 
ing the  process. 

37.  The  process  of  elimination  by  addition  or  subtraction  is 
effected  by  so  transforming  the  given  equations  that  the  unknown 
number  to  be  eliminated  appears  in  two  equations  with  coefficients 
which  differ,  if  at  all,  only  in  sign.  •  Then,  either  by  adding  or  sub- 
tracting the  corresponding  members  of  the  transformed  equations, 
the  terms  containing  this  unknown  may  be  made  to  disappear,  and 
the  resulting  derived  equation  will  be  free  from  this  unknown 
number.  After  solving  this  last  equation  for  the  single  unknown 
appearing  in  it,  the  value  of  the  other  unknown  may  be  found, 
either  by  substitution  or  by  repeating  the  process  above. 

Ex.  1.    Solve  the  system  of  linear  equations 

3a:4-4v  =  35,     (1))  t     ^-        o     ^ 
«      ^  .  /     )^i  r  I-    Given  System. 

6x-5y  =  U.     (2))  ^ 


SIMULTANEOUS  EQUATIONS  343 

To  eliminate  x  it  is  necessary  to  obtain  equations  equivalent  to  equations 

(1)  and  (2)  in  which  the  coefficients  of  a;  ditfer,  if  at  all,  only  in  sign. 
Multiplying  both  members  of  equation  (1)  by  2,  we  obtain  the  equivalent 

equation  6  a: +  81/ =70,  (3),  which,  if  taken  with  equation  (2),  forms  a 
System  II.  which  is  equivalent  to  the  given  System  I. 

0  a:  +  8  ?/  =  70,     (3)  ]  Equivalent 
y  II.  Derived 

6a; -57/ =  44.     (2) J  System. 

131/ =  26.     (4) 

By  subtracting  the  members  of  equation  (2)  from  the  corresponding 
members  of  ec^uation  (3),  the  terms  containing  x  disappear  and  the  derived 
equation  (4)  contains  ?/  only. 

Solving  (4),  we  obtain,     2/  =  2.     (5) 

By  Principle  (iii.)  §  35,  equation  (5)  taken  together  with  either  of  the 
original  equations,  say  (2),  forms  a  System  III.  which  is  equivalent  to  the 
given  system. 

7/  =    2,     (5)  \  Equivalent 

[■  III.  Derived 

6  a: -5?/ =  44.     (2))  System. 

By  substituting  the  value  2  for  y  in  one  of  the  given  equations,  say  (2), 
we  shall  obtain  an  equation  in  which  x  is  the  only  unknown  number. 
Solving  this  equation  we  shall  obtain  the  value  a:  =  9. 

Instead  of  obtaining  x  by  substituting  the  value  found  for  y  we  may 
transform  the  given  e(iuations  in  such  a  way  that  their  members  may  be 
combined  to  eliminate  ij,  Jis  follows : 

Multiplying  the  members  of  (1)  by  5, 

15a:  + 20y  =  175.  (())     .     .     .     .\  Equivalent 

Multiplying  the  members  of  (2)  by  4,  V  IV.        Derived 

24a: -20?/=  176.  (7)     ....  )  System. 

By  addition,      39  a:  =351 

Hence,  x  =9.  (8) 

a:  =  9  ) 
We  may  verify  the  solution     _  „'  r  V  substituting  these  values  in  the 

given  e<iuations,  obtaining  the.  identities  35  =  35  and  44  =  44. 

By  Principle  III,  Chap.  X.  §  28,  equation  (1)  in  System  I.  is  equivalent 
to  equation  (3)  in  System  II. 

By  Principle  (iii.)  §  35,  equation  (5),  obtained  from  equations  (3)  and 

(2)  of  System  II.,  taken  together  with  equation  (2),  forms  System  III. 
which   is   equivalent  to  the  original  System    I.     Hence   the  solution  of 


344  FIRST  COURSE  IN  ALGEBRA 

System    III.  must  be  the  solution  of  the  given  system  of  equations   (1) 
and  (2). 

By  a  similar  course  of  reasoning,  we  may  show  that  System  IV.  is 
equivalent  to  System  I. 

Hence,  the  solution  of  IV.,  which  is  the  same  as  that  of  III.,  is  the  solu- 
tion of  the  original  equations. 

The  results  of  the  algebraic  work  may 
be  illustrated  graphically  by  the  accompany- 
ing figure. 

The  graphs  are  numbered  to  correspond 
to  the  numbers  of  the  dillerent  equations. 
It  may  be  seen  that  the  point  marked  A  is 
the  intersection  of  the  graphs  (1)  and  (2), 
representing  the  given  equations  (1)  and  (2). 
It  is  also  located  by  the  graphs  (3)  and  (2), 
which  represent  the  ef^uations  of  System  II. ; 
by  the  graphs  (5)  and  (2),  which  represent 
the  equations  of  System  III. ;  by  the  graphs 
Pjq    jq  (6)  and  (7),  which  represent  the  equations 

of  System  IV.;  and  finally  by  the  graphs  (8) 
and  (5),  which  represent  the  solution,  a:  =  9,  y  =  2,  of  the  given  equations. 

38.  The  preceding  example  illustrates  the  following  rule  for 
elimination  by  addition  or  subtraction: 

First,  reduce  both  equations  to  the  farm  ax-\r  by  =c.  Multiply 
(yr  divide  both  members  of  the  equations,  if  necessary,  by  such  con- 
stants as  are  required  to  make  the  absolute  values  of  the  coefficients 
of  one  of  the  unknowns  equal  in  both  equations. 

Combine  the  corresponding  members  of  the  derived  equations  by 
addition  or  subtraction,  according  as  the  signs  of  the  coefficients  of 
the  unknown  numb&r  to  be  eliminated  are  unlike  or  like. 

Solve  the  resulting  equation  for  the  single  unknown  appearing 
in  it. 

To  find  the  remaining  unknown,  substitute  the  value  of  the  unknown 
just  found  for  the  sams  unknown  wherever  it  appears  in  one  of  the 
(yriginal  equations,  and  solve  the  derived  equation  for  the  single 
unknown  appearing  in  it ;  or  repeat  the  process  of  elimination  by 
addition  or  subtraction  for  the  remaining  unknown. 


SIMULTANEOUS  EQUATIONS  345 

Exercise  XVIL     4 

Solve  the  following  sets  of  simultaneous  equations,  eliminating- 
the  unknowns  by  the  process  of  addition  or  subtraction,  verifying 
all  results  : 

1.  9£c  +  7y  =  25,  15.  ^a;  -  eSj  =f\ 
dx-li/=  11.  d^x-cSj=f, 

2.  3  cc  +  4  ?/  =  47,  16.  r£c  +  57/  =  2  r.9, 

3  a!  +  2  3/  =  31.  sx  ■\-  ry  —  r^  -\-  s^, 

,3.  6a;  +  5^  =  23, 

4  a;  — 5?/=    7. 

4.  2a;+    hy—    4, 
4  a;—  10^  =  48. 

5.  7a;  +  y=16, 
^x  —  y=    4. 

6.  a; +2?/ =14, 
2a;  +  3?/ =  23. 

7.  5a;-8y  =  — 39, 
8a;—  5?/  =  0. 

8.  2a;—    ly  =  2, 
\i)x  —  2ly  =  -  18. 

9.  5a;  +  3y  — 90  =  0, 
2a;— 7?/  —  94  =  0. 

10.  3a;  +  4y  +  5  =  0. 
5a;  +  4?/+  3  =  0. 

11.  8a;-    9?/+    1=0, 
16a;  +  273/-  17  =0. 

12.  3  (a;  +  3/)  =  57, 
5(a;-y)  =  15. 

13.  ax+hy=  {a  —  hf, 
ax  —  by  =  a^  —  IP'. 

14.  X—    y  —  m  —  n^ 
mx  —  ny  =  1  (m^  —  n^). 


17. 

Fa;  +  rn^y  =  0, 

kx  +  my  =  k  +  m. 

18. 

X  -{-  ay  +  a^  =  Oy 

x+by+b^  =  0. 

19. 

(p  +  q)x  -(p-  q)y  =  3, 

{p-q)x-Y  {p  +  q)y  =  3. 

20. 

0.8  a; +  0.13/ =  0.19, 

0.6  a; +  0.93/ =  0.39. 

21. 

0.5  a;  +  0.4?/ =  0.13, 

0.7a;  +  0.33/ =  0.13. 

22. 

0.3  a; +  0.23/=    9.5, 

0.2  a;  +  0.33/=  10.5. 

23. 

-?=--. 

.+   1=17. 

24. 

|+3y  =  15, 

|  +  4a;  =  37. 

25. 

-  +¥  =  T 

¥-'=¥■ 

346  FIRST  COURSE  IN  ALGEBRA 

26.^+     5^    =9,  28.^+i^  =  3, 

y+9      g—  2^  a;  —  y  —  9  ^ 

10  3  *  ic  +  y  — 9        * 

2,.     £±-2-    ^-±1  =  0. 
4  5 


General  Solution  of  a  System  of  Two  Consistent, 

Independent,  Linear  Equations  Containing^ 

Two  Unknown  Numbers 

39.  A  system  of  two  consistent,  independent,  linear  equations, 
containing  two  unknowns,  has  one  and  only  one  solution. 

This  may  be  shown  by  solving  the  following  set  of  simultaneous 
equations  : 

«ia:  +  %  =  c„(l)|j    Given  System. 
«««  +  %  =  C2,  (2)  )  ^ 

In  these  equations  a^  6i,  Ci,  etc.,  are  read  "a  sub-one,"  "b  sub- 
one,"  "  c  sub-one,"  etc. 

The  subscripts  i  and  2  are  used  for  convenience  to  indicate  that 
the  letters  to  which  they  are  attached  are  found  in  either  the  first 
or  the  second  equation,  respectively. 

By  applying  the  principles  governing  the  derivation  of  equivalent 
equations,  it  may  be  seen  that  equations  (1)  and  (2)  may  be  taken 
as  representing  any  system  of  two  linear  equations  containing  two 
unknowns,  x  and  ?/;  a^  bi,  Ci,  «2,  ^2,  and  c^  being  known  numbers 
the  values  of  which  do  not  depend  upon  the  values  of  x  and  ^. 

We  may  apply  the  method  of  elimination  by  subtraction  as 
follows  : 


EUminating  y 

aJ3^x  +  hja^y  z=z  c.^b^ 

Eliminating  z 

a^a^  4-  b^a^  =  c^n^ 
a^a^x  +  b^a^y  =  c^a^ 
{b^a^  -  6,«i)7/  =  ci«2  - 

(ajb^  -  agijx  =  C162  -  C261  (3) 
a^b^-aj)^^ 

-«2^ 

1(6) 


SIMULTANEOUS   EQUATIONS  347 

By  Principle  (i.),  §  26,  and  Principle  (iii.),  §  35,  the  equations  (3)  and 
(4)  taken  together,  or  either  of  the  equations  (3)  or  (4),  taken  together  with 
one  of  the  given  equations  (1)  or  (2)  constitute  a  system  equivalent  to  the 
given  system. 

The  single  solution  of  equations  (3)  and  (4),  that  is,  equations  (5)  and 
(6)  taken  together,  is  accordingly  the  solution  of,  and  the  only  solution  of, 
the  given  system. 

It  may  be  seen  that  the  given  equations  (1)  and  (2)  are  equivalent,  pro- 
vided that 

either  (a^b^  —  a^hj)  =  0  and  (c^h^  —  c^h^)  =  0,  from  (3), 
or         («i&2  ~  <*2^i)  =  0  and  (cia^  —  c^a-^  =  0,  from  (4). 

It  m»y  also  be  seen  that  the  given  equations  are  inconsistent,  provided 
that  (a^ftg  —  ^2^1 )  =  ^^  '"^^^^  (^A  —  ^2^i)  ^  ^»  fi'om  (3), 

or        (a^ftg  —  «2^i)  ==  ^  ^"*^^  (^1^2  —  ^2^i)  ^  0,  from  (4). 

It  follows  that  the  condition  that  the  given  equations  shall  be  independent 
and  consistent  is  that  a^h^  —  a^h^  i^  (). 

Systems  of  Linear  ^Equations  Containing  Three 
Or  More  Unknowns 

40.  It  may  be  shown  that  a  system  of  three  or  more  consistent, 
independent,  linear  equations  has  in  general  a  single  definite  solu- 
tion, provided  that  the  number  of  equations  is  equal  to  the  number 
of  unknowns  appearing  in  them. 

The  solution  may  be  obtained  by  applying  the  methods  already 
shown  for  systems  containing  two  unknowns. 

41.  To  obtain  the  solution  of  a  system  of  three  consistent,  inde- 
pendent, linear  equations  containing  three  unknowns,  all  of  the 
unknowns  appearing  in  each  of  the  given  equations,  we  may  proceed 
as  follows  : 

Using  any  two  of  the  given  equations,  eliminate  one  of  the  un- 
knowns ;  then,  using  one  of  these  same  equations  with  the  remaining 
equation  of  the  system,  eliminate  the  same  unknown  as  before. 

Two  derived  equations  will  thus  he  obtained  which,  taken  together 
with  one  of  the  original  equations,  will  form  a  system  equivalent  to 
the  given  system. 

Solve  these  two  derived  equations,  for  the  two  unknowns  ajjpearing 
in  them,  by  the  methods  previously  shown;  substitute  the  values  of  the 


348  FIRST  COURSE   IN   ALGEBRA 

unknowns  thus  found  for  tke.se  unknowns  in  one  of  the  original  equa- 
tions to  obtain  ths  value  of  tJie  third  unknown. 

42.  Whenever  a  system  consists  of  four  equations  containing 
four  unknowns,  by  elimination  we  may  obtain  :  first,  a  system  of 
three  equations  containing  three  unknowns ;  then  from  this,  a 
system  of  two  equations  contjiining  two  unknowns ;  and  finally, 
a  single  equation  containing  a  single  unknown. 

The  value  of  the  single  unknown  obtained  by  solving  this  last 
equation  may  be  substituted  in  one  of  the  equations  containing 
two  unknowns  to  obtain  the  value  of  a  second  unknown  ;  substitut- 
ing the  values  of  these  two  unknowns  in  one  of  the  equations 
containing  three  unknowns,  we  may  find  the  value  of  a  third  un- 
known ;  the  value  of  the  fourth  unknown  may  be  obtained  by 
substituting  the  values  of  the  three  unknowns  for  these  unknowns 
in  one  of  the  equations  containing  four  unknowns. 

Graphical  Record  of  tlie  Process  of  Solution 

43.  The  follo^ving  device  will  be  found  to  be  helpful  in  planning 
and  keeping  record  of  the  different  steps  taken  when  solving  a 
system  of  simultaneous  equations. 

Since,  by  transposing  all  of  the  terms  of  a  given  equation  to  the 
first  member,  we  may  derive  an  equivalent  equation  in  which  the 
second  member  is  zero,  it  is  seen  that  the  first  member  of  the  equa- 
tion thus  transformed  is  a  function  of  the  unknowns  appearing  in 
it.  It  follows  that  any  equation  containing  a  single  unknown,  x^ 
may  be  represented  by  the  notation /(cc)  =  0,  read  "function  of  x 
equal  to  zero " ;  an  equation  containing  two  unknowns,  x  and  ?/, 
by  the  notation  /(a;,  y)  =  0,  read  "  function  of  x  and  y  equal  to 
zero  "  ;  and  an  equation  containing  three  unknowns,  x,  y^  and  z,  by 
the  notation  f{x^  y,  z)  =  0,  read  "  function  of  x,  y,  and  z  equal  to 
zero." 

Diff*erent  equations  may  be  denoted  by  subscripts. 

Thus, ^ (a;,  y,  z)  =  0,  read  "function  one  of  x,  y,  and  z  equal  to 
zero," ^(ic,  y,  z)  =  0,  read  "function  two  of  x,  y,  and  z  equal  to 
zero,"  and  /gCa;,  y,  z)  =  0,  read  "  function  three  of  x,  y,  and  z  equal 
to  zero,"  represent  three  different  equations  which  may  be  referred 
to  as  equations  (1),  (2),  and  (3),  each  containing  x,  y,  and  z. 


SIMULTANEOUS  EQUATIONS  B49 

The  letters  written  within  each  parenthesis  must  in  every  case  be 
the  same  as  the  different  unknown  letters  appearing  in  the  equa- 
tion numbered  to  correspond  to  the  subscript  of  the  symbol. 

By  this  notation  our  attention  is  directed  simply  to  the  fact  that 
certain  equations  contain  particular  unknowns,  and  nothing  is  indi- 
cated as  to  the  forms  of  the  equations  in  which  the  unknowns  are 
found. 

If,  using  equations  (1)  and  (2),  one  of  the  unknowns,  say  y,  is 
eliminated,  the  derived  equation  will  contain  the  two  unknowns 
£c  and  ;s.  This  derived  equation  may  be  represented  by  the  symbol 
f^  (x,  z)  =  0.  The  derivation  of  equation  (4)  from  equations  (1) 
and  (2)  may  be  suggested  as  follows  : 

A  fc  y.  z)  =  0 — -^^/4  fe  z)  =  0. 

If,  during  the  process  of  elimination,  the  members  of  equations 
(1)  and  (2)  are  multiplied  by  a  and  b  respectively,  this  may  be 
indicated  by  placing  a  and  b  on  the  leading  lines,  as  shown  above. 

A  second  equation  (5)  containing  the  same  unknowns,  x  and  z, 
which  may  be  represented  by  the  symbol  /^(x,  z)  =  0,  may  be 
derived  by  using  either  equations  (1)  and  (3)  or  equations  (2) 
and  (3). 

If  equations  (1)  and  (3)  are  used,  the  derivation  may  be  sug- 
gested as  follows: 

/i  (a^,  y,z)=o  ^^^^ 

If  equations  (4)  and  (5)  are  used  to  eliminate  z  to  obtain  a  single 
equation  (6)  in  which  x  alone  appears,  we  may  suggest  the  deriva- 
tion as  follows: 

A  (x,  z)  =  0 — 2:==^^/,  (x)  =  0. 

The  forms  of  the  equations  may  be  such  that  it  is  unnecessary 
to  use  multipliers  represented  by  a,  b,  c,  d,  e,  and/,  which  are  shown 
on  the  leading  lines. 

It  will  be  seen  that  the  solution  of  the  given  system  of  three 


350  FIRST  COURSE  IN  ALGEBRA 

equations,  (1),  (2),  and  (3),  containing  three  unknowns,  is  made  to 
depend  upon  the  solution  of  a  system  of  two  equations,  (4)  and  (5), 
in  which  two  unknowns,  x  and  z^  are  found.  The  sohition  of  this 
system  in  turn  depends  upon  the  solution  of  a  single  equation  (6) 
in  which  a  single  unknown,  a;,  appears. 

44.  It  will  be  found  that  this  device  will  enable  the  student  to 
plan  the  solution  of  a  system  of  equations  intelligently,  and  to  carry 
out  the  work  systematically,  with  the  advantage  that  at  any  stage . 
of  the  process  the  record  will  show  the  reason  for  each  step  taken, 
and  serve  also  as  a  guide  for  such  additional  steps  as  are  necessary 
to  complete  the  solution. 

45.  To  illustrate  the  use  of  this  graphical  record,  we  will  solve 
a  system  of  three  simultaneous,  independent,  linear  equations  con- 
taining three  unknowns. 

Ex.  1.    Solve  the  following  system  of  simultaneous  equations  : 
4a:  +  3y  +  93  =  53,  (l)-\ 
11  a:  -  2  y  +  8  2  =  75,  (2)  [■  Given  System. 
6a:  +     y +  55;  =  47.  (3)) 
An  examination  of  the  equations  shows  that  it  will  be  convenient  to 
select  y  as  the  unknown  to  be  eliminated  first. 

The  unknown,  y,  may  be  eliminated  by  using  the  first  and  third  equa- 
tions and  also  the  second  and  third  equations. 

Two  equations  which  may  be  called  equations  (4)  and  (5)  will  thus  be 
obtained,  in  each  of  which  the  two  unknowns  x  and  z  will  be  found. 

After  the  equations  (4)  and  (5)  are  obtained  the  remaining  steps  of  the 
process  of  solution  may  be  determined. 

We  begin  the  record  of  the  process  of  solution  by  writing  the  symbols 
representing  the  set  of  given  equations  j  that  is,  the  equations  numbered 
1,  2,  and  3. 

The  elimination  of  y  from  equations  (1)  and  (3),  and  also  from  equations 
(2)  and  (3),  to  produce  the  derived  equations  numbered  (4)  and  (5),  may 
be  suggested  as  follows : 

/i  (^,  2/,  2:)  =  0  ^..^^^^^^^^^ 

h  (^,  v^  ^)  =  0  -^^^^^^^  ^'''  "^^  =  ^ 

/3  (X,  7/,  .)  =  0  ^:^^=^J^  {X,  .)  =  0 

The  multipliers,  3  and  2,  which  are  used  to  multiply  the  members  of 
equation  (3)  in  preparation  for  the  elimination  of  ij  to  obtain  equations  (4) 
and  (5)  respectively,  are  written  on  the  leading  lines,  as  shown  above. 


SIMULTANEOUS  EQUATIONS  351 


Equations  (4)  and  (5)  may  be  derived  as  follows  : 


(2)  Unaltered,  liar— 2?/+  82=  75 

(3)  Modified. 
Members 

mult,  by  2.  l2x+2y-\-lOz=:  94 

By  addition,  23a:  +18^=     169.  (5) 


(1)  Unaltered.     4x+3y+  9z=  53 

(3)  Modified. 
Members 
mult,  by  3.    18a:+3?/+15;3=141 

By  subtraction,  14a:       +  6s=  88, 
Hence,  7x       +  Sz=  44.(4) 

The  solution  of  the  given  system  of  three  equations  containing  three 
unknowns,  x,  y,  and  2,  is  thus  made  to  depend  upon  the  solution  of  a  derived 
systeiu  consisting  of  two  equations  containing  two  unknowns,  x  and  z. 
7a:  +    3z=    44,     (4)  ) 
23a:+  18;3=  169.     (5)  [ 

It  should  be  understood  that  this  derived  system  is  not  equivalent  to  the 
given  system  consisting  of  three  equations. 

However,  the  derived  system  consisting  of  equations  (4)  and  (5),  taken 
together  witli  one  of  the  given  equations,  will  form  a  system  of  equations 
which  is  equivalent  to  the  given  system  of  three  equations. 

Using  equations  (4)  and  (5),  we  may  eliminate  z  by  multiplying  the 
members  of  equation  (4)  by  6  and  then  combining  the  members  of  the 
resulting  ec^uation  by  adtlition  with  the  members  of  equation  (5).  A  sixth 
equation,  containing  a  single  luiknown,  x,  will  thus  be  obtained,  and  this 
may  be  solved  for  x. 

The  graphical  record  of  the  process  of  solution  may  now  be  completed  by 
indicating  the  elimination  of  z  by  using  equations  (4)  and  (5)  to  obtain 
equation  (6),  as  follows  : 

/3  (X,  y.  z)  =  O-^^S^/5  C^,  ■^)  =  0 _^=-^/,  (x)  =  0 

The  process  of  the  derivation  of  equation  (6)  may  be  carried  out  as  follows  : 

(4)  Modified. 

Mem.  mult,  by  6.  42  a:  +  1 8  z  =  264 

(5)  Unaltered.  23a:  +  I8z=  169 
By  subtraction,  19a:  =95. 
Hence,                          x  =      5.     (6) 

By  substituting  the  value  of  x,  which  is  5,  for  x  in  either  of  the  equations 

(4)  or  (5)  and  solving  the  resulting  equation,  we  shall  obtain  the  value  of 
z,  as  follows : 

Using  (4)  we  have,  7  •  5  +  3  rj  =  44.     Hence  z  =  3. 


352 


FIRST  COURSE  IN  ALGEBRA 


To  obtain  the  value  of  y  we  may  select  an  equation  containing  y,  say 
equation  (3),  and  substitute  for  x  and  z  the  values  5  and  3  respectively, 
and  solve  the  resulting  equation  as  follows ; 

6  •  5  +  1/  +  5  •  3  =  47. 

Hence,  y  =  2. 

The  solution  of  the  given  system  of  simultaneous  equations  is  the  set  of 

x  =  5, 
values  y 

z 


LI     \Jl       Olil 


From  the  process  of  derivation,  it  will  be  seen  that  roots  have  neither 
been  gained  nor  lost  during  the  transformation.  Hence,  these  values  con- 
stitute the  only  solution  of  the  given  set  of  simultaneous  equations. 

Substituting  in  the  original  equations  for  z,  y,  and  z  the  values  5,  2, 
and  3  respectively,  we  obtiiin  the  following  numerical  identities  :  53  =  53, 
(1) ;  75  =  75,  (2)  ;  and  47  =  47,  (3). 

Ex.  2.  Solve  the  following  system  of  four  independent,  simultaneous, 
linear  equations  containing  four  unknowns,  x,  y,  z,  and  w. 

3x+    y-2z-\-     w=    1,  (1)  ' 

2x-4y  +5m7=    5,  (2) 

3y  +  2«-4«;=  16,  (3) 

4a;-     2/  +  3Z  =  35.  (4) 

The  accompanying  graphical  outline  will  serve  to  indicate  the  different 
steps  of  the  process. 
/i(x,y,z,«;)  =  Ov 
fi(.x,  y,     w)  =  0  N.      /gCar,  y,  w)  =  0^^^ 


Given  System. 


h{x,y,z,    )  =  0 


/6(^»!/»*0  =  <^- 


— Vg(x,u')=0 


Equations  (5)  and  (6)  may  be  derived  as  follows: 

(3)  Modified. 
(1)  Unaltered.  3x+  y-2  2+   w=    1 


(3)  Unaltered.  3i/+2;^-4M;=  16 


(5)  By  addit'n,  3  a:+4  y        -3  iw  =  17. 


Mem.  mult. 

by  3. 
(4)  Modified. 
Mem.  mult.    8x 

by  2.         

(6)  By  sub- 
traction 


./,(«;)  =  0. 


9  2/+6;s-12Mfci48 


2  i/-l-6  z  =70 


d>x—\\y        -\-l2tv=22 


SIMULTANEOUS   EQUATIONS 


35B 


It  will  be  seen  that,  when  %  is  eliminated  by  using  equations  (1),  (8), 
and  (4),  the  derived  equations  (5)  and  (6)  contain  the  unknowns  a;,  ^, 
and  w. 

Hence,  equation  (2)  may  be  carried  over  unaltered  as  the  remaining 
ec][uation  necessary  to  complete  the  set  of  three  equations  containing  three 
unknowns,  a:,  y,  and  w. 

The  solution  of  the  given  system  of  equations  is  thus  made  to  depend 
upon  the  solution  of  the  three  following  equations  : 

3a:  +  4^-  Zw=.  17,  (5)") 
2a:-  4^+  5w=  5,  (2)  [■ 
8a:-  ll2/  +  12ii?  =  22.  (6)) 


Equations  (7)  and  (8)  may  be  derived  as  follows ; 

(2)  Modified 


(5)Unaltered3a;+47/-3t(;=17 
(2)Unaltered2a:— 4i/+5w=  5 


By  addition,    bx    4-2r6-22.(7) 


Mem.  mult,  by  11. 
(6)  Modified 
Mem.  mult. by  4. 


22a;-447/+55w=55 

32a: -44^/  + 48m; =88 


(8)  By  subtraction,    lOx 


lw-2>Z. 


The  solution  of  the  given  system  of  three  equations  has  now  been  made 
to  depend  upon  the  solution  of  the  following  system  of  two  equations : 
5a:  +  2w;=22,     (7)  > 
10a;-7ry  =  33.     (8)) 


Equation  (9)  may  be  derived  as  follows : 
(7)  Modified 


Mem.  mult,  by  2. 
(8)  Unaltered, 
By  subtraction 


10a;  +  4«;  =  44 

10a;-  7wr=33 
11m;=  U 
m;=    1. 


(9) 

By  substituting  the  value  1  for  w  in  equation  (7)  or  equation  (8)  we  may 
obtain  the  value  of  x. 

Using  (7),  we  have  5  a;  +  2  •  1  =  22.     Hence  a:  =  4. 

By  substituting  the  values  4  and  1  for  x  and  w  respectively  in  equations 
(5),  (2),  or  (6),  we  may  find  the  value  of  y. 

Using  (5),  we  have  3-4  +  42/-3-l  =  17.     Hence  ?/  =  2. 

The  value  of  z  may  be  found  from  equations  (1),  (3),  or  (4),  by  substi- 
tuting the  values  4,  2,  and  1  for  x,  y,  and  w  respectively. 

Using  (3),  and  substituting  the  values  for  y  and  iw,  we  have 

3-2  +  22;- 4-1  =  16.     Hence  z  =  7. 
23 


354  FIRST  COURSE   IN  ALGEBRA 

The  solution  of  tbe  <j;iven  set  of  simultaneous  equations  is  thus  found  to 
be  the  following  set  of  values  : 

a-  =  4/ 
!/  =  2. 

2  =  7, 
V)  =  1 . 

Froui  the  nature  of  the  derivation  of  the  successive  equations  it  may  be 
seen  tliat  solutions  have  neither  been  gained  nor  lost.  Hence  the  solution 
found  is  the  only  solution  of  the  given  system  of  simultaneous  equations. 

Substituting  these  values  for  x,  y,  z,  and  «?,  respectively,  in  the  given 
equations,  we  obtain  the  following  numerical  identities  : 

1  =  1,(1);  5  =  5,  (2) ;  16  =  16,  (3)  ;  and  35  =  35,  (4). 

Exercise  XVII.     5 

Solve  the  following  systems  of  simultaneous  equations,  verifying 
all  solutions  : 

6.  a;  +  ^  +  i;  =  19, 
y  =  2a;-    3, 
z=    y  —  10. 

7.  3ic+  4?/  +  52;  =  — 68, 
2x+  y  =  —  2, 
4?/  —     z  =  —  14. 

8.  X—     i/  —  2z  =  —  4:, 
x  —  2y—    z  =  —  4:, 

2x—    y  —    z  =  0' 

9.  6  a;  —  y  =  3  z  —  S6, 
ij  —  3z  =  3x  —  S9y 
z-2x=Sy+    2. 

10.  2x—3y  =  ^z-2\, 
2y  —  Sz  =  4:X—  14, 
2z  —  Sx  =  4:y—  10. 

5.  5x  +  4:y+Sz  =  S5,  11.  2x  —  3y  +  2 z  =  j^^, 

4tX+Sy+2z  =  25,  3a;-2y+  Sz  =  h 

Sx+2y-     z=15.  2x+3y-2z  =  U' 


1. 

3x+2y  = 

13, 

Sy+2z  = 

8, 

Sz+  2x  = 

9. 

2. 

a:  —  y  =  Sy 

y-^  =  h 

z  +  x  =  Q. 

3. 

,-1  =  6. 

4. 

2x+  5y- 

Sz  = 

=  23, 

3ic-f  2y  + 

*?   •r  — 

=  41, 

5x-4:y-\-  Qz  = 

=  35. 

SIMULTANEOUS  EQUATIONS  355 


12.  2x-i-  3y  +  Az-^  5  =  0, 
2x  +  3y-4:Z+  6  =  0, 
2x  —  3y  +  4;:  +  7  =  0. 

13.  lOaj  — 8y+     2;  =  40, 

x+     y  +  2z=    5, 
3a;—     z  +  6     =    0. 

14.  0.3  a;  +  0.5^  =  0.8, 
0.4  a;  +  0.7  2^  =  1.8, 
0.1 7/  +  0.1  2;  =  0.3. 

15.  0.1  a;  +  0.3?/=  1.9, 
0.2  a;  4-  0.4  5;  =  3.2, 
O.oy  +  0.1;2  =  3.1. 

16.  X  +  y  =  c, 
I/  +  z  =  b, 
z  +  X  =  a. 

17.  X  +  y  =  a  +  by 
y  +  z  =  b  +  c, 
z  -\-  x  =  c  +  a. 

18.  x+  3y  =  a, 
y-^3z  =  b, 
z  -\-  3  a;  =  c. 

19.  X  +  y  —  z  =  a, 
x  —  y  +  z  =  b, 
z  +  y  —  x  =  c. 

20.  bx  +  ay  -{■  cz  =  a, 
ex  4-  by  i-  az  =  b, 
ax  +  cy  +  bz  =  c. 


Systems  of  Fractional  Equations  Solved  like  Equations  of 
the  First  Degree 

46.  Certain  systems  of  fractional  equations  which  are  linear  with 
reference  to  the  reciprocals  of  the  unknowns  may  be  solved  without 
clearing  of  fractions  before  eliminating  the  unknowns. 


21. 

,  2x  —  3y=  6, 

4.y-6z=7y 

2z'-3u  =  8, 

Au  —  5x=d. 

22. 

2x-y  =    8, 
3y-z  =13, 

4:Z  —   W=1Q, 

5w—x  =13. 

23. 

3x  +  Ay-2z  = 

:20, 

2x—  1  y  +  5u  = 

=  -9, 

8x+2z  +  3u  = 

--21, 

2y—3z  +  4:U  = 

--  17. 

24. 

lx+    y  ^  Az  = 

:      W, 

a;  +    w  —    y  = 

■-    0, 

2z—    w  -\-3y  = 

=   15, 

3y-lx-2z  = 

:  3W  - 

25. 

x  +  y  +  z=3, 

y  -{-  z  +  u  =A, 

z  +  ti  +  x  =  5j 

u  +  X  +  y  =  Q. 

26. 

X  +  y  +  z  +  u  = 

12, 

x+y+z+v= 

14, 

x  +  y  +  u-\-v  = 

16, 

X  -\-  z  -\-  u  +  V  = 

18, 

y  +  z  +  u  +  V  = 

20. 

356  FIRST  COURSE   IN  ALGEBRA 

Ex.  1.  Solve  the  system  of  fractional  equations 


15      18      „      „J 

-  +  -  =  9,     (1) 

X      y 
20        6 

-I. 

Given  System 

r-  -y  =  '-  ^'\ 

Instead  of  clearing  of  fractions  before  eliminating  either  of  the  unknowns, 
in  which  case  we  should  introduce  into  the  ei^uations  terms  containing  both 

X  and  y,  we  will  eliminate  the  reciprocals  of  the  unknowns,  -  and  - ,  and 

then  from  the  derived  equations  obtain  values  for  x  and  y. 


Multiplying  members  of 
equation  (2)  by  3. 

^-'1=    6. 
X         y 

(3) 

(1)  Unaltered. 

15  +  1?=    9. 
X^  y 

(1) 

By  addition, 

Z5       =15. 

X 

Hence, 

75         r 

The  value  of  y  may  be  obtained  by  substituting  5  for  x  in  one  of  the 
given  equations  and  solving  the  resulting  equation  for  y,  or  by  repeating 
the  process  of  elimination  as  follows : 
Multiplying  members  of 

equation  (1)  by  4. 

Multiplying  members  of 

equation  (2)  by  3. 

By  subtraction, 
Hence, 

Thus,  it  appears  that  the  following  set  of  values  is  the  solution  of  the 
given  system  of  simultaneous  equations: 


X       y 

^-15=    6. 

X        y 

(4) 
(6) 

?2  =  30. 
2'  =30  =  3- 

x  =  5,) 
y  =  3.i 


y 

Substituting  these  values  in  the  original  equations,  we  obtain  the  numeri- 
cal identities  9  =  9,  (1),  and  2  =  2,  (2). 

From  the  process  of  derivation  it  may  be  seen  that  solutions  have  been 
neither  gained  nor  lost.  Hence  the  set  of  values  found  is  the  only  solution 
of  the  given  system  of  equations. 

47.  When  solving  systems  of  fractional  equations  it  should  be 
kept  in  mind  that  during  the  process  of  clearing  the  fractional 


SIMULTANEOUS  EQUATIONS  357 

equations  of  fractions  no  solutions  will  be  lost,  but  it  may  happen 
that  extra  solutions  will  be  introduced.  (Compare  with  Chapter 
X.  §  30,  also  Chapter  XVI.  §  4.) 


Exercise  XVII.     6 

Solve  the  following  systems  of  equations  which  are  fractional  with 
reference  to  the  unknowns,  verifying  all  results  obtained  : 


1. 

1       1 

X      y 

a:      y 

2. 

9       10 

X        y 

3. 

X      y      A: 

X     y      ^ 

4. 

4         7       5 

6_14__1^ 
X       y            2 

5. 

X       y 

X     y 

6. 

3      4_  7 
X      y      24* 

8      6        1 
X      y      12* 

7. 

2a; +3^-^' 

11         10 
4a;"^93^"  3 

8. 

1           1         9 
bx      12y      2' 

15a;  '     Gy~^' 

9. 

L+^r'"' 

'   +   '=20. 
5x      6y 

10. 

— +  — -i, 
ax      by      c 

4  +  9-   1   . 
ax      by      2d 

11. 

a    ,   b'' 

—  +  —  =  c, 
x       y 

X      y 

12. 

11  a;-- =  3, 

10a;~-  =  4. 

858 


13.  -  +  - 
X      y 

X      y 

14.^  +  ^  = 

X      y 

a      h  _d 
X      y~  c 


FIRST 

COURSE 

IN 

ALGEBRA 

1, 

21. 

i*r^' 

3. 

\*\-'. 

c 

y    - 

Hint.   From  the  sums  of  the 

correspoiidiug  members  of  all  of 

3  J  the  e(iuutioii8,  subtract  the  cor- 

1*^*  "    +  z 3  =  0,  responding  members  of  each  of 

X         6  y 


t-\*\- 


the  equations  in  turn,  and  solve. 


„.     3^7        43  22- 

»"•    S   +  ^  =  20' 

2y-6^  =  -I^. 
•'  10 

bx      ay  ' 

"^     y  23. 

18.  hx—  by  =  4:xy, 
Qy  +  Gx  =  5xy. 
Hint.   Divide   both  members 
of  each  of  the  equations  by  xy. 


19. 


X 

4 

+  1 

+ 

y 

5 
-2 

X 

5 
+  1 

— 

y 

3 
-2 

X 

2 

+  2 
3 

+ 

y 

3 

+  3 
2 

=  9, 

=  2. 

-4  = 

0, 

+  4  = 

0. 

24. 


1       2 

+  3  = 

X      y 

=  0, 

1       ^       . 

=  0, 

x;      a: 

=  0. 

X      y      a 

-  H —  =  7> 

y     z      b 

2;      ic      c 

X      y      z 

1 

2' 

20.-^  +  ^^-4  =  0.  l-\  +  \-\' 

X+-2       1/  +  S  X      y      z       i 


PROBLEMS 

X      y      s        ' 

27. 

XII 

X      y      z 

,:=-'• 

X      y      z 

ZX 

.  +  .  =  "• 

359 


12       8  Hint.    Write  the  first  equation 

.     26.  -  +  -  +  -  =  20,  111 

^       y       z  .     in  the  form  -  +  -  =  -. 

2       3        1  y      X      a 

— I 1 —  =  17^  Write    the    others   in   similar 

^       y       '^  forms. 

X      y      z 

Problems  Involving  Simultaneous  Equations 

48.  Whenever  the  unknown  numbers  of  a  problem  are  obtained 
by  solving  algebraic  conditional  equations,  it  is  necessary  that  the 
number  of  independent  consistent  equations  be  equal  to  the  number 
of  unknowns  whose  values  are  to  be  found. 

49.  It  is  often  a  matter  of  choice  whether  a  particular  problem 
shall  be  solved  by  using  a  single  equation  containing  one  unknown, 
or  a  system  of  two  or  more  independent  equations  containing  two 
or  more  unknowns. 

Exercise  XVIL     7 

1.  Separate  101  into  two  such  parts  that  2/5  of  the  greater  shall  exceed 
2/3  of  the  less  by  2. 

Let  X  stand  for  the  greater  of  the  two  numbers  into  which  101  is  sep- 
arated, and  let  y  represent  the  less  number. 

By  the  conditions  of  the  problem  we  obtain  the  conditional  equations 
a:+     2/ =101, 
2a:      %y 

T  "  T  -  ^• 

The  solution  of  these  equations  is  found  to  be  a;  =  65,  which  is  the  greater 
number,  and  y  =  36,  which  is  the  less  number. 

Tliese  numbers  are  found  to  satisfy  the  conditions  of  the  given  problem. 

2.  Find  a  fraction  such  that,  if  7  be  added  to  both  numerator  and  denomi- 


360  FIRST  COURSE   IN  ALGEBRA 

naU)r,  its  value  becomes  4/5,  and  if  2  be  subtracted  from  botli  numerjitor 
and  denominator,  its  value  becomes  1/2. 

Let  X  represent  the  numerator  and  y  the  denominator  of  the  fraction. 
Then  by  the  conditions  of  the  given  problem  we  have 

ar  +  7  .    4 


and 


y  +  7-6' 
g-2      1 

y-2~2 


The  solution  of  these  two  ecjuations  is  found  to  be  a:  =  5  (which  is  the 
numerator  of  the  fraction),  and  y  =  8  (which  is  the  denominator). 

Accordingly  the  required  fraction  is  5/8. 

This  fraction  will  be  found  to  satisfy  the  conditions  of  the  problem  as 
stated. 

3.  The  difference  between  two  numbers  is  5  and  their  sura  is  29.  Find 
the  numbers. 

4.  Two  numbers  are  to  each  other  in  the  ratio  of  7  to  9,  and  if  50  be 
subtracted  from  each  of  the  numbers  the  remainders  will  be  to  each  other 
as  1  is  to  2.     Find  the  numbers. 

5.  Separate  109  into  two  parts  such  that  3/8  of  the  greater  part  shall 
exceed  4/9  of  the  less  by  4. 

6.  If  three  times  the  greater  of  two  numbers  be  divided  by  the  less,  the 
quotient  is  3  and  the  remainder  is  15;  and  if  four  times  the  less  be  divided 
by  the  greater,  the  quotient  is  3  and  the  remainder  is  14.  What  are  the 
numbers  ? 

7.  What  is  that  fraction  which  equals  1/5  when  1  is  added  to  the  numer- 
ator, and  equals  1/6  when  1  is  added  to  the  denominator  ? 

8.  Find  a  fraction  which  is  equal  to  1  /5  when  its  numerator  and  denom- 
inator are  each  diminished  by  2,  and  is  equal  to  1  /  3  when  its  terms  are 
increased  by  3. 

9.  If  1  is  added  to  the  numerator  of  a  certain  fraction  its  value  becomes 
5/7,  and  if  1  is  added  to  the  denominator  the  value  becomes  3/5.  What 
is  the  fraction? 

10.  Find  a  fraction  such  that  if  1  be  added  to  both  numerator  and  de- 
nominator the  value  becomes  .1/2,  while  if  1  be  subtracted  from  both 
numerator  and  denominator  the  value  becomes  7/16, 

11.  If  3  be  added  to  both  numerator  and  denominator  of  a  certain  frac- 
tion its  value  becomes  7/9;  if  3  be  subtracted  from  both  numerator  and 
denominator  its  value  becomes  1/3.     Find  the  fraction. 

12.  If  the  numerator  of  a  certain  fraction  be  multiplied  by  2  and  its 
denominator  be  increased  by  5,  the  value  of  the  fraction  becomes  1/2  ;  if 


PROBLEMS  361 

the  denominator  be  multiplied  by  2  and  the  numerator  be  increased  by  18, 
the  value  of  the  fraction  becomes  unity.  Find  the  numerator  and  denom- 
inator of  the  fraction. 

13.  The  numerator  and  denominator  of  a  certain  proper  fraction  each 
consists  of  the  same  two  figures  whose  sum  is  9,  written  in  different  orders. 
If  the  value  of  the  fraction  be  3/8,  find  the  numerator  and  denominator. 

14.  The  numerator  and  denominator  of  a  certain  improper  fraction  each 
consists  of  the  same  two  figures  whose  sum  is  6,  written  in  different  orders. 
If  the  value  of  the  fraction  be  7/4,  find  the  numerator  and  denominator. 

15.  Separate  1(K)  into  three  parts  such  that  if  the  second  part  be  divided 
by  the  first  the  quotient  is  3  and  the  remainder  2;  and  if  the  third  be  divided 
by  the  second  the  quotient  is  3  and  the  remainder  is  1. 

16.  A  immber  expressed  by  two  figures  is  equal  to  7  times  the  sum  of 
its  figures.  If  27  be  subtracted  from  the  number,  the  figures  in  tens'  and 
units'  places  are  interchanged.     Find  the  number, 

17.  Separate  the  two  numbers  75  and  70  into  two  parts  each,  such  that 
the  sum  of  one  part  of  the  fii-st  and  one  part  of  the  second  shall  equal  100, 
and  the  difference  of  the  remaining  parts  shall  equal  25. 

18.  Separate  the  two  numbers  60  and  50  into  two  parts  each,  such  that 
the  sum  of  one  part  of  the  first  and  one  part  of  the  second  shall  equal  75, 
and  the  difference  of  the  remaining  parts  shall  equal  5. 

19.  Separate  a  into  two  parts  such  that  1/mth  of  the  greater  part  shall 
exceed  l/wtii  of  the  less  by  b. 

20.  A  number  is  composed  of  two  figures  whose  sum  is  12.  If  the 
figures  in  tens'  and  units'  places  are  interchanged  the  number  is  increased 
by  18.     Find  the  number. 

21.  A  farmer  bought  100  acres  of  land  for  $3304.  If  part  of  it  cost 
him  $50  an  acre  and  the  remainder  $18  an  acre,  find  the  number  of  acres 
bought  at  each  price. 

22.  If  five  pounds  of  sugar  and  ten  pounds  of  coffee  together  cost  $3.80, 
and  at  the  same  price  ten  pounds  of  sugar  and  five  pounds  of  coffee  cost 
$2.35,  what  is  the  price  of  each  per  pound  ? 

23.  A  man  invested  $5000,  a  part  at  5  per  cent  and  the  remainder  at 
4  per  cent  interest.  If  the  annual  income  from  both  investments  was  $235, 
what  were  the  separate  amounts  invested  ? 

24.  There  are  two  pumps  drawing  water  from  a  tank.  When  the  first 
works  three  hours  and  the  second  five  hours,  1350  cubic  feet  of  water  are 
withdrawn.  When  the  first  works  four  hours  and  the  second  three  hours, 
1250  cubic  feet  of  water  are  withdrawn.  How  many^cubic  feet  of  water 
can  each  pump  discharge  in  one  hour  ? 

25.  A  plumber  and  his  helper  together  receive  $4.80.     The  plumber 


362  FIRST  COURSE  IN  ALGEBRA 

works  5  hours  and  the  helper  6  hours.  At  another  ti)iie  the  plumber  works 
8  hours  anil  the  helper  9^  hours,  and  they  receive  $7.65.  What  are  the 
wages  of  each  per  hour  ? 

26  Three  men  and  three  boys  can  do  in  4  days  a  certain  amount  of 
work  whicti  can  be  done  in  6  days  by  one  man  and  5  boys.  How  long 
would  it  require  one  man  alone,  or  one.  boy  alone,  to  do  the  work  ? 

27.  A  certain  piece  of  work  can  be  completed  by  3  men  and  6  boys  in 
2  days.  At  another  time  it  is  observed  that  an  equal  amount  of  work  is 
performed  in  3  days  by  1  man  and  8  boys.  Find  the  length  of  time  re- 
quired for  one  man  alone  or  for  one  boy  alone  to  do  the  given  amount  of 
work. 

28.  Two  persons,  A  and  B,  can  complete  a  certain  amount  of  work  in  I 
days;  they  work  together  m  days,  when  A  stops;  B  finishes  it  in  n  days. 
Find  the  time  each  would  require  to  do  it  alone. 

29.  A  steamer  makes  a  trip  of  70  miles  up  a  river  and  down  again  in 
24  hours,  allowing  5  hours  for  taking  on  a  cargo.  It  is  observed  that  it 
requires  the  same  time  to  go  2^  miles  up  the  river  as  7  miles  down  the 
river.  Find  the  number  of  hours  required  for  the  up  trip  and  for  the 
down  trip  respectively. 

30.  A  train  ran  a  certain  distance  at  a  uniform  rate.  If  the  rate  had 
been  increased  by  4  miles  an  hour  the  journey  would  have  required  16 
minutes  less,  but  if  the  rate  had  been  diminished  by  4  miles  an  hour  the 
journey  would  have  required  20  minutes  more.  Find  the  length  of  the 
journey  and  the  rate  of  the  train  in  miles  per  hour. 

31.  A  steamer  runs  a  miles  up  a  river  and  back  again  in  t  hours.  It  is  ob- 
served that  it  requires  the  same  time  to  go  b  miles  with  the  stream  as  it 
does  to  go  c  miles  against  it.  Find  expressions  for  the  number  of  hours 
required  for  the  up  and  down  trips  respectively,  and  also  for  the  velocity  of 
the  stream  in  miles  per  hour. 

32.  Three  trains  start  for  a  certain  city,  the  second  h  hours  after  the 
first  and  the  third  k  hours  after  the  first.  The  second  and  third  run  at  the 
rates  of  a  and  b  miles  an  hour  respectively.  If  all  three  arrive  together, 
find  an  expression  for  the  distance  and  for  the  rate  of  the  first  train  in  miles 
per  hour. 

33.  A  marksman  fires  at  a  target  600  yards  distant.  He  hears  the 
bullet  strike  4  seconds  after  he  fires.  An  observer,  standing  525  yards  from 
the  target  and  300  yards  from  the  marksman,  hears  the  bullet  strike  3 
seconds  after  he  hears  the  report  of  the  rifle.  Find  the  velocity  of  the  sound 
in  yards  per  second  and  also  the  velocity  of  the  bullet  in  yards  per  second, 
supposing  each  to  be  uniform. 

34.  Having  given  two  alloys  of  the  following  composition  :  A,  composed 


PROBLEMS  363 

of  4  jiaits  (by  weight)  of  gold  and  3  of  silver ;  B,  2  parts  of  gold  and  7  of 
silver  ;  how  many  ounces  of  each  must  be  taken  to  obtain  6  ounces  of  an 
alloy  containing  equal  amounts  (by  weight)  of  gold  and  silver  ? 

35.  Two  alloys,  A  and  B,  contain :  A,  2  parts  (by  weight)  of  tin  and  9 
parts  of  copper  ;  B,  7  parts  of  tin  and  3  parts  of  copper.  To  obtain  1000 
pounds  of  alloy  containing  (by  weight)  5  parts  of  tin  and  16  parts  of  copper, 
how  many  pounds  of  each  must  be  taken  and  melted  together  ? 

36.  A  bar  of  metal  contains  20.625  per  cent  pure  silver,  ami  a  second 
bar  12.25  per  cent.  How  many  ounces  of  each  bar  must  be  used  if,  when, 
the  parts  taken  are  melted  together,  a  new.  bar  weighing  50  ounces  is  ob- 
tained, of  which  15  per  cent  is  pure  silver  ? 

37.  A  and  B  run  two  quarter-mile  races.  In  the  first  race  A  gives  B  a 
start  of  2  seconds  and  beats  him  by  20  yards.  In  the  second  race  A  gives 
B  a  start  of  6  yards  and  beats  him  by  4  seconds.  Find  the  rates  of  A  and  B 
in  yards  per  second. 

38.  A  and  B  run  a  race  of  500  yards.  In  the  first  trial  A  gives  B  a 
start  of  7  yards  and  wins  by  10  seconds.  In  the  second  trial  A  gives  B  a 
start  of  56  yards  and  wins  by  2  seconds.  Find  the  rates  of  A  and  B  in 
yards  per  second. 

39.  In  a  race  of  one  hundred  yards  A  beats  B  by  \  of  a  second.  In  the 
second  trial  A  give's  B  a  start  of  3  yards,  and  B  wins  by  1 J^  yards.  Find 
the  time  required  for  A  and  B  each  to  run  100  yards. 

40.  A  and  B  run  a  race  of  440  3'^ard8.  In  the  first  trial  A  gives  B  a 
start  of  65  yards  and  wins  by  20  seconds.  In  the  second  trial  A  gives  B  a 
start  of  34  seconds  and  B  wins  by  8  yards.  Find  the  rates  of  A  and  B  in 
yards  per  second. 

Solve  the  following-  problems,  employing"  equations  con- 
taining three  or  more  unknowns: 

41.  If  270  is  added  to  a  certain  number  of  three  figures  the  figures  in 
tens'  and  hundreds'  places  are  interchanged.  When  198  is  subtracted  from 
the  number  the  figures  in  hundreds'  and  units'  places  are  interchanged. 
The  figure  in  hundreds'  place  is  twice  that  in  units'  place.     Find  the  number. 

42.  A  number  is  expressed  by  three  figures  whose  sum  is  10.  The  sum 
of  the  figures  in  hundreds'  and  units'  places  is  less  by  4  than  the  figure  in 
tens'  place,  and  if  the  figures  in  units'  and  tens'  places  are  interchanged  the 
resulting  number  is  less  by  54  than  the  original  number.  Find  the  original 
number. 

43.  Find  three  numbers  such  that  the  sum  of  the  reciprocals  of  the  first 
and  second  is  1/2  ;  of  the  second  and  third,  1/3  ;  and  of  the  third  and  first, 
1/4. 


364  FIRST  COURSE  IN  ALGEBRA 

44.  Separate  400  into  4  parts  such  that  if  the  first  part  be  increased  by 
9,  the  second  diminished  by  9,  the  third  multiplied  by  9,  and  the  fourth 
divided  by  9,  the  results  will  all  be  equal. 

45.  A  and  B  together  can  do  a  certain  piece  of  work  in  7^  days,  A  and 
C  in  6  days.  All  three  work  together  for  2  days,  when  A  stops  and  B  and 
C  finish  the  work  in  2^  days.  How  long  would  it  require  each  man  alone 
to  do  the  work  ? 

46.  A  and  B  can  do  a  piece  of  work  in  r  days,  A  and  C  can  do  the  same 
work  in  s  days,  and  B  and  C  can  do  it  in  t  days.  Find  in  how  many  days 
each  can  do  the  work  alone. 

47.  In  a  mile  race  A  can  beat  B  by  60  yards  and  can  beat  C  by  230  yards. 
By  how  much  can  B  beat  C  ? 

Represent  the  rates  of  A,  B,  and  C  in  yards  per  second  by  a,  6,  and  c 

respectively. 

The  time  required  for  A  to  run  one  mile  or  1760  yards  is  1760/a  seconds. 

Since  A  beats  B  by  60  yards,  B  in  the  same  time  runs  1700  yards.     The 

time  required  by  B  is  1700/6  seconds. 

Accordingly  we  have  the  conditional  equation 

1760  _  1700 

a     ~     h     '  ^^ 

Similarly,  since  A  can  beat  C  by  230  yards,  we  have  the  conditional 

1760      1530     ,^^ 
equation  = (2) 

From  (1)  and  (2)  we  obtain  — j—  =  — — ,  (3),  from  which  we  find  that 

C's  rate  and  B's  rate  must  satisfy  the  conditional  equation  c  =  ^^  b.     (4) 

If  B  and  C  run  a  mile  race  the  time  required  for  B  is  1760/  b  seconds. 

Let  X  represent  the  number  of  yards  by  which  B  beats  C.  Then  the 
time  required  for  C  to  run  1760  —  x  yards  is  (1760  —  x)/c  seconds. 

Accordingly,  we  have  the  conditional  equation 
1760  1760  -  X  .^. 
-^  =  —c ^'> 

Substituting  for  c  in  (5)  the  expression  ^j^  b  from  (4),  and  solving  the 
resulting  equation  for  x,  we  obtain  x  =  176,  which  is  equal  to  the  number 
of  yards  by  which  B  beats  C. 

This  value  is  found  to  satisfy  the  conditions  of  the  given  problem. 

48.  In  a  race  of  500  yards  A  can  beat  B  by  20  yards,  and  C  by  30  yards. 
By  how  many  yards  can  B  beat  C  ? 

49.  Having  given  3  bars  of  metal,  the  first  containing  (by  weight)  6  parts 
of  gold,  2  parts  of  silver,  and  1  part  of  lead  ;  the  second,  3  parts  of  gold,  4 


PROBLEMS  365 

parts  of  silver,  and  2  parts  of  lead ;  the  third,  1  part  of  gold,  3  parts  of  silver, 
and  5  parts  of  lead ;  find  how  many  ounces  of  each  must  be  taken  to  obtain 
12  ounces  of  an  alloy  containing  equal  amounts  (by  weight)  of  gold,  silver, 
and  lead. 

50.  Of  three  bars  of  metal,  the  first  contains  13  parts  (by  weight)  of 
silver,  5  parts  of  copper,  and  2  parts  of  tin  ;  the  second,  35  parts  of  silver, 
4  parts  of  copper,  and  1  part  of  tin  ;  the  third,  8  parts  of  silver,  7  parts  of 
copper,  and  5  parts  of  tin.  How  many  ounces  of  each  bar  must  be  used  if 
when  the  parts  taken  are  melted  together  a  bar  is  obtained  which  weighs 
10  ounces,  of  which  5  ounces  are  silver,  3  ounces  are  copper,  and  2  ounces 
are  tin  ? 


866  FIRST  COURSE  IN  ALGEBRA 


CHAPTER  XVIII 
EVOLUTION 

1.  A  POWER  has  been  defined  as  the  product  of  two  or  more  equal 
factors.     (See  Chap.  V.  §  25.) 

With  respect  to  the  power,  each  of  the  equal  factors  is  called  a 
root. 

According  as  there  are  two,  three,  four,  or  n  equal  factors,  each  is 
called  a  square  root,  cube  root,  fourth  root,  or  wth  root. 

E.  g.    Since  3*  =  3  x  3  =  9,  3  is  a  square  root  of  9. 
Again,  since  (—  3)'-'  =  (—  3)(—  3)  =  +  9,  —  3  is  also  a  square  root  of  9. 
Also,  since  2*  =  2x2x2  =  8,  2  is  a  cube  root  of  8. 
We  shall  see  later  that  there  are  in  all  three  expressions  whose  cubes  are 
8,  hence  there  are  three  different  cube  roots  of  8. 

2.  The  radical  or  root  sigrn  /y/  written  before  a  number  is  a 
sign  of  operation  which  is  commonly  used  to  denote  that  a  root  is  to 
be  taken.  This  symbol  is  an  abnormal  form  of  the  initial  letter  r, 
from  the  Latin  radix  meaning  root. 

3.  The  number  or  expression  whose  root  is  required  is  called  the 
raclicand. 

E.  g.  The  expression  y'g  means  that  the  square  root  of  9  is  to  be  taken. 
The  number  9  is  called  the  radicand. 

4.  A  number  called  the  index  of  the  root  is  written  before  and 
directly  above  the  radical  sign  to  indicate  which  root  is  required. 

E.  g.  The  symbols  ^^  ^^  ^y/,  ^,  denote  that  the  second,  third,  fourth, 
and  wth  roots,  respectively,  of  the  numbers  or  expressions  before  which  they 
are  placed  are  to  be  taken. 

In  case  no  index  is  written,  the  index  2  is  understood,  so  that  \/  indicates 
that  the  square  root  is  required. 


EVOLUTION  367 

5.  The  radical  sign  affects  only  the  factor  before  which  it  is  placed. 
Accordingly  if  the  radicand  consists  of  more  than  one  factor  or  more 
than  one  term,  it  must  either  be  enclosed  in  parentheses,  or  an 
overline  or  vinculum  joined  to  the  radical  sign  must  be  used  to 
denote  that  the  root  of  the  expression  underneath  is  to  be  taken  as 
a  whole. 

E.  g.  Observe  that  the  expression  <y/4  x  9  means  that  the  square  root 
of  4  is  to  be  taken  and  the  result  is  to  be  multiplied  by  9.  Hence,  since 
the  square  root  of  4  is  either  +  2  or  —  2,  it  follows  that  the  expression 
/y/4  X  9  represents  either  +  18  or  —  18. 

The  expression  is  considered  to  be  arranged  in  better  form  if  written 
as  9.Y/4. 

If  the  square  root  of  the  product  of  4  and  9  is  required,  we  may  write 
either  ^^(4  x  9)  or  \/4:  x  9.  Since  the  square  root  of  the  product  4x9, 
which  is  36,  is  either  +  6  or  —  6,  either  of  these  expressions  may  be  taken 
as  meaning  +  6  and  also  —  6. 

The  expression  \/'S6  +  64  means  that  the  square  root  of  the  sum  of  36 
and  64,  which  is  100,  is  to  be  taken. 

Hence  /\/36  +  64  means  either  +  10  or  —  10. 

6.  According  as  their  indices  are  equal  or  unequal,  two  roots  are 
said  to  be  like  or  unlike  without  regard  to  the  equality  or  in- 
equality of  the  radicands. 

E.  g.   The  expressions  \/x  and  v^  are  like  roots, 
while  -v/m  and  \/m  are  unlike  roots. 

7.  A  root  is  said  to  be  even  or  odd  according  as  its  index  is 
even  or  odd. 

E.  g.   The  expressions  \/a,  \h^  v  7,    v  10  denote  even  roots, 

while  Vc,  V^'  ^'^'^y  VlO  denote  odd  roots. 


8.  A  number  or  expression  which  can  be  expressed  either  as  an 
integer  or  as  a  fraction  of  which  the  terms  are  finite  integers,  with- 
out using  an  indicated  root,  is  said  to  be  rational  or  commen- 
surable. 

In  the  contrary  case  it  is  said  to  be  Irrational  or  incommen- 
surable. 


368  FIRST  COURSE  IN  ALGEBRA 


E.  g.  Since  \/l6  can  be  expressed  as  either  +  4  or  —  4  it  follows  that 
^li5  is  a  rational  number ;  while  \/2,  which  cannot  be  expressed  either  as 
a  whole  number  or  as  a  fraction  whose  terms  are  finite  integers,  is  an  irra- 
tional or  incommensurable  number. 

9.  A  number  or  expression  is  said  to  be  an  nth  power  if  its  nth. 
root  is  a  rational  number  or  expression. 

E.  g.  The  number  36  is  a  square  because  either  of  its  square  roots,  +  6 
or  —  6,  is  a  rational  number. 

The  number  27  is  a  cube  because  one  of  its  cube  roots  is  a  rational 
numljer  3. 

The  number  17  is  not  an  nth  power,  because  a  rational  number  cannot  be 
found  which  is  its  nth  root. 

10.  As  a  formal  definition  of  a  root  (letting  n  represent  a 
positive  whole  number)  we  have 

(^a)~  =  a. 
Or  \/a  is  one  of  the  7i  equal  factors  of  a. 

General  Principles  Governing  Root  Extraction 
Number  of  Roots. 

(i)  A  positive  number  has  at  least  two  even  roots  which  are  eqiial 
in  absolute  value  but  opposite  in  sign. 

We  may  use  (±  3)'^  =  9  as  a  convenient  abbreviation  for  the  two 
identities  (+  3)^=  9  and  (-3)'=  9. 

Since  (+  3)''  =  9  and  (-  3)'  =  9  it  follows  that  V^  is  +  3  and 
also  —  3.  It  is  convenient  to  show  that  Vl)  has  two  values  by- 
writing  a/9  =  ±  3.  It  should  be  understood  that  a/9  =  ±  3  is  not 
an  identity,  but  is  simply  a  convenient  abbreviation  for  two  identi- 
ties a/9  =  +  3  and  a/9  =  —  3. 

We  may  write  Va'^  =  ±  «  as  a  convenient  abbreviation  for  the 
two  identities  Va^  =  +  a  and  Va^^  =  —  a. 

Similarly  v^^^"  =  ±  b  is  to  be  understood  as  meaning  that 
VP^  =  +  ^  and  v^  =  -L- 

(ii.)  There  exists  at  least  one  odd  root  of  any  positive  or  negative 
number  which  has  the  same  sign  as  the  number  itself. 


EVOLUTION  869 

For  siDce   (+  cf^^  =  +  c^^+i,  we  have  '^""^Z?^     =  +  c.     (1). 

Also  since  (-  c)'"'^^  =  -  c=^»+\  we  have  "*ty_  ^.2«+i  ^  _  c.     (2). 

Changing  the  signs  of  both  members  of  (1),  we  obtain 

_^"V^^       =_e.  (3). 

It  follows  from  (2)  and  (3)  that  ''*1^- c-^'-^  =  -  '"I^X?^^ 

From  the  reasoning  above  it  appears  that  : 

(iii.)  J^hr  the  operation  of  finding  an  odd  root  of  a  negative  number 
may  be  substituted  that  of  finding  a  like  odd  root  of  a  positive  number 
having  the  same  absolute  value,  provided  that  a  negative  sign  is  pre- 
fixed to  the  result. 

E.  g.    i^r^=_^27  =  -3. 

11.  The  principal  root  of  a  positive  number  is  its  single  posi- 
tive root. 

E.  g.    Tlie  principal  square  root  of  4  is  2;  of  9  is  3  ;  of  16  is  4;  etc. 

12.  The  principal  odd  root  of  a  negative  number  is  its  single 
negative  root. 

E.  g.  The  principal  cube  root  of  —  27  is  —  3  ;  the  principal  fifth  root  of 
—  32  is  -  2 ;  etc. 

13.  The  radical  sign  will  be  used  to  denote  the  principal  root 
only,  unless  the  contrary  is  expressly  stated. 

That  is,  ^^'  =  \a\. 

In  order  that  the  signs  of  operation  +  and  —  may  be  applied  to 
rational  and  irrational  numbers  without  exception,  we  shall  under- 
stand that  whenever  a  term  of  an  algebraic  expression  is  affected  by 
a  radical  sign  the  principal  root  of  the  radicand  is  to  be  taken ;  and 
this  root  is  to  be  combined  with  the  other  terms  of  the  expression 
by  addition  or  subtraction,  according  as  the  radical  sign  is  preceded 
by  a  -f  or  a  —  sign. 

It  should  be  observed,  however,  that  the  root  of  a  monomial  which 
is  not  a  term  of  an  algebraic  expression  may,  according  to  circum- 
stances, be  positive  or  negative. 

Thus,  V25  =  ±  5  ;   V4  =  ±  2. 

24 


370  FIRST  COURSE  IN  ALGEBRA 

It  should  be  observed  that  although  'y/25  =  ±  5  and  \/4  =  ±  2,  the 
expression  ^^25  +  ^^4  =  5  +  2=7;  also,  the  expression  \/2b  —  'y/4  = 
5-2  =  3. 

The  expression  V^5  +  ^/J  does  not  mean  ±  5  ±  2,  vvliich  is  either  +  7 
or  —  7. 

Whenever  an  even  root  is  required,  and  we  have  means  for  know- 
ing that  a  given  radicand  is  an  even  power  of  a  negative  number, 
it  is  necessary  that  a  negative  number  be  taken  as  the  root. 

Thus,  if  in  a  calculation  a  for  some  reason  is  regarded  as  being  negative, 
then  if  a  is  raised  to  the  second  power  the  result  a*  must  be  considered  as 
the  s(juare  of  a  negative  number,  not  as  the  square  of  a  positive  number. 

Accordingly,  in  such  a  case  we  would  have  ^/a^  =  —  |  a  |. 

In  particular,  VC-  1)(-  1)  =  V{-  I)^  =  -  1. 

14.  Since  (+  5)^  =  +  25  and  (—  5)^  =  +  25,  it  may  be  seen 
that  —  25  cannot  be  obtained  by  multiplying  together  two  like 
foctors  which  are  either  both  positive  or  both  negative.  Accord- 
ingly it  is  impossible  to  express  V—  25  either  as  a  positive  or  a 
negative  number. 

Representing  any  even  number  by  2n,  it  may  be  seen  that 
(±  aY"  =  +  a''^".  It  is  therefore  impossible  to  express  V—  a'^" 
either  as  a  positive  or  a  negative  number. 

In  Chap.  XXI  we  shall  deal  with  indicated  even  roots  of  nega- 
tive numbers  which  are  commonly  called  imaginary  numbers. 

To  distinguish  them  from  these  so-called  imaginary  numbers,  all 
other  numbers,  such  as  those  with  which  we  have  previously  dealt, 
are  called  real  numbers. 

The  following  principles  and  proofs  apply  to  real  numbers  only. 

15.  The  operation  of  finding  any  required  power  of  a  given 
number  or  expression  is  called  involution,  and  that  of  finding 
any  required  root  of  a  given  number  or  expression  is  called 
evolution. 

16.  It  may  be  shown,  if  we  extend  somewhat  our  idea  of  number, 
that  there  are  two  square  roots,  three  cube  roots,  four  fourth  roots, 
etc.,  and  n  nth  roots  of  a  given  number  or  expression. 


EVOLUTION  371 

Roots  of  Monomials 

Principles  Governing  Operations  with 
Radical  Symbols 

17.  As  fundamental  to  root  extraction  we  have  the  following 
Principle :  Like  principal  roots  of  equal  numbers  or  expressions 

are  equal. 

18.  In  the  statements  of  the  following  principles  the  principal  root 
is  the  root  meant  in  each  case,  and  for  convenience  of  proof  we  shall 
restrict  the  radicand  a  to  represent  a  positive  whole  number  the 
value  of  which  cannot  become  infinitely  great. 

19.  Root  of  a  Power. 

(i.)  The  exponent  of  the  rth  root  of  a  given  base  a^  is  found  by 
dividing  the  exponent  rx  of  the  power  by  the  index  r  of  the  indicated 
root. 

That  is,  A/a****  =  «»*, 

By  (iii.)  Chap.  VII.  §  1,  a"'"  =  (a")'". 

Hence  ^~^r  =  ^J^^ny^ 

Or  -v/rt^  =  a«. 

E.g.    ^n^  =  a\ 

20.  Root  of  a  Product. 

(ii.)  The  rth  root  of  a  product  of  two  or  more  factors  is  equal  to 
the  product  of  the  rth  roots  of  the  given  factors. 

That  is,  ^Jabcd  .  .  .  )  =  '^Ti^hVc^d    •  •  . 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

We  will  establish  this  principle  for  a  radicand  consisting  of  two  factors, 
and  by  similar  reasoning  the  result  may  be  extended  to  include  three  or 
more  factors. 

Let  jp  represent  the  positive  value  of  '\/a\/b. 

That  is,  V  =  Va\^b. 

Raising  both  members  to  the  rth  power,  we  have 

Or,  2>''  =  ^'^^- 


372  FIRST  COURSE  IN  ALGEBRA 

Accordingly  p  is  one  of  the  rth  roots  of  the  product  aft,  and  since  it  is 
real  and  positive,  it  must  be  the  principal  root. 
Hence,  ,^  =  ^a^. 

Similarly,  ^abcd =  ^^yb^/c^/d 

Ex.  1.  ^49  X  81  =  ^^49  X  '^  =  7  X  9  =  (>3. 

Ex.  2.  v^-8a«6«  =  \/^^'^^^\/b^  =  -  2  ab\ 

Mental  Exercise  XVIII.  1 
Find  the  indicated  root  of  each  of  the  following  expressions  : 

1.  \^¥.    .  10.  ^?V?:  19.  v'G4aW^ 

2.  V^^^  11.   ^xYz'^  20.   ^-343««^iV^ 

3.  V^  12.  -v^ajiy^".  21.   V441  a^°^V. 

4.  V^"^  13.  Virt^  22.  ^1296  6V«6^«. 

5.  V?^  14.   V257i^  23.   ^-32^^°</i^ 

6.  \^¥F:  15.   V^=^8^  24.   ^1024^%^V^ 

7.  Va*h'c\  16.   V^27a«6^  25.  '^64a«6^. 

8.  ^^^^f?.  17.   ^^2lW.  26.   Va^^'b^^'iT. 

9.  \^xyz'\  18.   ^16a^  27.    \/¥VW. 
21.     Root  of  a  Root. 

(iii.)  TV/^*  wM  roo^  ^y'  M^  7'^A  ro(9^  o/  «wy  radicand  is  equal  to  the 
nrth  root  of  the  given  radicand. 

That  is,  V^^  =  "-V^. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time. ) 

If  we  consider  principal  roots  only,  v^a  must  have  a  positive  value. 
The  nth  root  of  the  positive  value  oi^^a  must  itself  have  a  positive  value. 
If  this  nth  root  be  represented  by  ^,  we  have  p  =  y  v^a. 
Raising  both  members  of  the  equation  to  the  nth  power,  we  have  7?"=  -v/a. 
Raising  both  members  to  the  rth  power,  we  have  (^")''  =  a. 
Hence,  j^"'-  =  a.     (See  (iii.)  Chap.  VII.  §  1.) 

Accordingly  p  is  one  of  the  wrth  roots  of  a,  and  since  it  is  positive,  it 
must  be  the  principal  root. 


EVOLUTION  373 

Hence,  p  =  ^a  . 

Consequently  V  l/«  =  V^«  • 

Ex.  1.  ^^^64  =  -^64  Ex.  2.  Wela^P  =  ^2%«6i2 

=  2.  =2a62. 

22.    Power  of  a  Boot. 

(iv.)    The  pth  power  of  the  rth  root  of  any  radicand  is  equal  to  the 
rth  root  of  the  pth  power  of  the  given  radicand. 
That  is,         {.^/ay  =  V^  . 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Since  a  is  positive,  we  have 

a  =  '^^p. 

Accordingly,  ^a  =  \  "^ av. 

=  ^^. 
We  have  ^op  =  ^  ^J~aP. 


Hence,  ^  =  V^a^. 

Raising  both  members  to  the  ^th  power, 

Or,  (^a'')  =  ^^. 

Ex.  3.    ('^4)8  =  ^48  =  A^64  =  8. 

Exercise  XVIII.     2 
Write  the  following  roots  of  roots  as  rational  expressions  : 

1.  yl^'a^.  7.  yj~¥¥^^\  13.  {V2^xY\ 

2.  ^V(N}\  8.  ^~\/a^^¥^.  14.  {\/bMcy\ 

3.  V^'V^64^.  9.  V^^^^V^.  15.  {^/J^y. 

I.  yl'WW^.  10.  {^/'ahy.  le.  (v'^^v^)^". 

5.  VVIe^v.       11.  (^^^^)^  17.  (A/a«^)-". 


6.  S^¥^^^\        12.  {^/2xyy.  18.   (^i«y;5''''*)''*+^ 


374  FIRST  COURSE  IN  ALGEBRA 

23.    Root  of  a  Quotient. 

(v.)  The  rth  root  of  a  quotient  is  eqiml  to  the  quotient  obtained,  by 
dividing  the  rth  root  of  the  dividend  by  the  rth  r^^ot  of  the  divisor . 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time. ) 

If  a  and  h  be  both  positive,  h  being  difterent  from  zero,  and  we  represent 
the  value  of  the  positive  fraction  a/  b  by /and  of  the  whole  number  b  by  w, 
we  have,  applying 

Hence  ^^b  =  (^^'fj  {Tb, 

Therefore,  V^^  =  (\/ 1)  Vb. 

Dividing  both  members  by  \/b,  we  have 


Or 

Ex.  1. 


</6~      ^/b 

/25_  V25_5 
V  36  ~  V36  ~  6  ' 


Mental  Exercise  XVIII.    3 
Simplify  each  of  the  following  roots  of  quotients  : 


64c?V 


EVOLUTION  375 

Square  Roots  of  Polynomials 

24.  We  have  shown  in  Chap.  XII.  §§  18,  19,  a  method  for  ob- 
taining by  inspection  the  equal  factors,  and  hence  the  square  root, 
of  a  trinomial  which  is  a  square. 

We  will  now  present  another  method. 

Representing  any  trinomial  sijuare  by  a^  +  2  ah  +  ^^  we  may 
obtain  the  square  root  as  follows : 

Arranged  according  to  descending  powers  of  a,  we  may  write 

r/  +  2«6  +  ^'  =  a'  +  (2a  +  h)h.  (1) 

We  may  obtain  the  first  term  in  the  required  square  root  by 
taking  the  square  root,  a,  of  the  first  term,  a^. 

Subtracting  the  square  of  a  from  the  given  trinomial,  we  obtain 
as  a  jivi^t  remainder  2ab  +  0'^,  which  may  be  written  in  the  form 
(2  a  +  0)b. 

The  second  term  of  the  required  square  root  may  be  found  by 
dividing  the  first  term  2  a/j  of  the  first  remainder  2  ab  +  P,  ar- 
ranged according  to  descending  powers  of  a,  by  a  tfial  divisor  2  a, 
which  is  formed  by  multiplying  by  2  the  part  of  the  root  already 
found. 

From  (1)  it  appears  that  by  increasing  2  a  hy  b  we  may  obtain 
a  complete  divisor,  2a  +  b,  which  when  multiplied  by  b  will  produce 
the  terms  2ab  +  b^  of  the  first  remainder. 

Subtracting  the  product  of  2  a  -f  ^  and  b,  that  is,  2  ab  +  b\  fi-om 
the  first  remainder,  2  a^  +  b^,  we  obtain  zero  as  a  second  remainder. 

The  steps  of  the  process  are  shown  below: 

Given  Expression      Square  Root 


First  term  of  root,     \/a^  =  a. 


i^  +  2ab  +  h^  I  a-\-b 


Trial  divisor,  2  a. 

4-  2  a6  +  62 

Second  term  of  root,     2ah-^2a  =  b. 

Complete  divisor,     2a  -f  &. 

(2a  +  b)h        = 

2  a&  -f  fe2 

376  FIRST  COURSE   IN   ALGEBRA 

Ex.  1.    Find  the  square  root  of  9x<  —  42x-i/  +  491/*. 


First  term  of  root,    \/9x*     =3  x\ 
(3x=»)«  = 


Given  Expression  Square  Root 

9a:*  -  42  x'Y  +  -ly  y^  |3x--7j/8 
9x* 


Trial  divisor,    2(3x2)  =  Gx^*. 

-42xY  +  49i/« 

Second  term  of  i-oot,   —  42  x^y*  ■^6x^=- 

■7/. 

Complete  divisor,  6  x*  —  7  //*. 

(6x«-7i/»)(-72/»)  = 

-42xV  +  492/« 

(3) 


25.  "We  will  now  show  that  the  steps  of  the  process  may  be 
repeated  to  obtain  the  sc^uare  root  of  any  polynomial  square  which 
contains  more  than  three  terms. 

The  scjuare  of  a  polynomial  may  be  written  as  follows  : 

For  two  terms,  (a  +  Oy  =  a^  +  2  «/>  j  ,  . 

For  three  terms,        (a  -\-  0  -[■  c-y  =  a^  -\-2ab  +  2ac\ 

+     Ir  ■{■  2bc\{2) 

For  four  terms,  (a  +  6  +  c  +  </)*  =  a^  +  2  a^  +  2  ac  +  2  «c?^ 

+      b^  +  2hc  ■^2bd 
H-      c^  +  2cd 
+      ^. 

Factoring  the  groups  of  terms  which  appear  in  the  vertical 
columns,  the  identities  (1),  (2),  and  (3)  become  respectively: 

(a  +  by  =  a^+  (2a  +  b)b.  (1) 

(a  +  b  +  cf  =  a^  +  (2a  +  b)b+  (2  a  +  2b  +  c)c.  (2) 
(a  +  b  +  c-{-  d)''  =  a^+  (2a  +  b)b+l2a+  2b  +  c)c  + 

(2a  +  2b  +  2c  +  d)d.  (3) 

It  may  be  seen  that,  with  each  new  letter  added  on  the  left,  a 
new  group  of  terms  in  parentheses  is  added  on  the  right.  This  new 
group  consists  of  the  product  of  twice  the  sum  of  all  previous  letters 
plus  the  last  letter,  multiplied  by  the  last  letter. 

In  (1),  (2),  and  (3)  the  expressions  in  parentheses  are  in  each  case 
the  complete  divisors  used  in  the  extraction  of  the  square  root  of  a 
polynomial  at  the  successive  stages  of  the  process. 


EVOLUTION* 


377 


Ex.  2.    Find  the  square  root  of 

a2  +  2  a6  +  6=2  +  2  ac  4-  2  6c  +  c2  +  2  afZ  +  2  &rf  +  2  c(Z  +  ^2. 


rt^  = 


Given  Expression  Square  Root 

a^+2  ab-\-b^+2  ac+2  bc-{^^+2  ad+2  bd-\-2  cd+d^  \  a^h^-c+d 
a2 


2  X  rt  =  2  rt. 

2ab 

2rt6-^2«          =6. 

2a  +h. 

(2a  -{-b)b         = 

2  ab-\-b^ 

2(a  +  6)            =2a-\-2b. 

2ac 

2ac-^2a          =c. 

2a  -\-2b-\-c. 

(2a  +26  +  c>  = 

2  fiw+2  bc-\-(!'^ 

2{a-\-b-\-c)  =  2a  +  2b^ 

-2c. 

2  ad 

2ad-^2a     =d. 

2a-\-2b-\-2c-\-d. 

(2a-\-2b  +  2c-\-d)d  = 

2ad+2bU^2cd+d^ 

26.  The  method  employed  in  §§  24  and  25  for  extracting  the 
square  root  of  a  polynomial  may  be  stated  in  the  following  form 
as  a  rule. 

Rule  for  finding  the  principal  square  root  of  a  poly- 
nomial square. 

Write  the  given  polynomial  according  to  descending  or  ascending 
powers  of  some  letter  of  arrangement. 

Extract  the  square  root  of  the  first  term  and  write  the  result  as 
the  first  term  of  the  required  square  root. 

Subtract  the  square  of  the  first  term  of  the  root  from  the  given 
polynomial^  and  arrange  the  fir.Ht  remainder  according  to  the 
powers  of  the  letter  of  arrangement ^  and  in  the  same  order  as  before. 

Divide  the  first  term  of  the  remainder  by  twice  the  first  term  of  the 
root,  write  the  quotient  as  the  second  term  of  the  root,  and  add  it  also 
to  the  trial  divisor  to  form  the  com^plete  divisor. 

Subtract  from  the  first  remainder  the  product  of  the  complete 
divisor  multiplied  by  the  term  of  the  root  last  found,  and  arrange 
the  remainder,  if  there  be  one,  as  a  second  remainder. 

Repeat  the  process,  using  as  a  trial  divisor  at  each  stage  of  the 
work  twice  the  part  of  the  root  already  found. 


378 


FIRST  COURSE   IN  ALGEBRA 


Ex.  3.    Find  the  square  root  of  29a^ -  40a^+  16a«-  46a«+  4  +  49a*-  12a. 
Arrangetl  according  to  descending  powers  of  a  we  have 

Given  Elxpression  Square  Root 


\/l6a«=4rt». 
(4a«)2  = 


16a«-40a»+49a*-46a»+29a»-12a+4  |4a»-5a«+3a-2 
16a« 


2x4a»=    8a>. 

-4()a6 

-40a»-r8a?=-5a». 

8as-^a«. 

(8ii»^a»X-6o»)  = 

-40a«+25a* 

2(4o5  — 5a»)  =  8o» 
24a<-f8<j3  =  3a. 
8a»-10a»+3a. 
(8a»— 10aH-3«)3a  = 


10  a». 


+  24a* 

+  24«*-30rt»+  9a2 


2  (4  a»  —  5  a» -f- 3  a)  =  8fla— 10o» + 6a. 
—  IScs^Sa'^-  2. 
8a'«— 10a»-|-6a  — 2. 
(8  a>  — 10  o»  +  6a  — 2X — 2)  = 


16<t8  +  2()a2 
16a»+20a2-12a+4 


In  practice  the  student  should  obtain  the  successive  terms  of  the  root  by 
performing  the  divisions,  —  40  a^  -f  9  a*  =  —  5  a^,  etc.,  mentally. 

Exercise  XVIII.     4 

Find  the  square  roots  of  the  following  expressions  : 

1.  a*  +  2a^  —  a^  —  2a+  1. 

2.  Ux*  —  '2-kiiiP  +  25x^—l2x-\-4. 

3.  81a;*  + 54a^  +  81a^+ 24«+  Ifi. 

4.  9a;*  +  24a-^  +  22a;*  +  38a;*  +  41  x^  +  10a;  +  25. 

5.  256«  — 306*  — 206^*  + 96'+  12^> +  4. 

6.  36a'+ 48a6+  12ac  +  166'  + 8^6'  + c*. 

8.  a;*  +  3  +  -,-2a;'-^- 

a;*  x^ 

9.  4  -  4j/  +  13/  +  16/  +  17/  —  22/  -  24/. 
10.   4a;*"+ 12a;«"  +  29ar'"  + 30af  +  25. 

Cube  Roots  of  Polynomials 

27.   A  process  for  finding  the  cube  root  of  a  polynomial  which  is 
a  cube  may  be  developed  as  follows  : 

We  know  that     (a  +  6)8  =  «'  +  3  a^O  +  3ab^  +  b\ 


EVOLUTION 


379 


Hence,    ^a^-h'S  a^b+'d  ab^+b^  =  \^a^+(S  a'+^  ab+b^)b  =a  +  b.  (1) 

It  may  be  seen  that  the  first  term  a  of  the  required  cube  root  is 
the  cube  root  of  the  first  term  of  the  given  expression  arranged 
according  to  descending  powers  of  a. 

Subtracting  the  cube  of  the  first  term  a  of  the  root  from  the 
given  expression,  we  obtain  as  &>  first  remainder  Sa'^b  +  Sab^  +  b^. 

Dividing  the  first  term  of  this  first  remainder,  arranged  according 
to  descending  powers  of  a,  by  a  trial  divisor,  3  a^,  which  is  obtained 
by  taking  three  times  the  square  of  the  first  term  of  the  root  already 
found,  we  obtain  the  second  term,  6,  of  the  root. 

To  obtain  a  complete  divisor  we  may,  as  indicated  in  (1),  add  to 
the  trial  divisor,  3  a%  the  term  3  ab  which  is  three  times  the  product 
obtained  by  multiplying  the  first  term  a  of  the  root  by  the  second 
term  b,  and  add  also  the  square  of  the  second  term  b  of  the  root, 
that  is,  b^. 

If  the  complete  divisor  thus  obtained,  3  a^  +  3  ab  +  b"^,  be  multi- 
plied by  b  and  the  product  be  subtracted  fi:om  the  first  remainder, 
3  a^b  4-  Sab^  +  b^y  we  have  zero  as  a  second  i-emainder^  and  accord- 
ingly the  process  stops  here  and  the  required  cube  root  \^  a  ■\-  b. 

The  different  steps  of  the  process  are  shown  below  : 

Cube  Foot 

\    a  +  h. 


First  term  of  root,     V  «*  =  «• 

0 
c 

Given  Expression 

i»  +  3  a26  +  3  ab'^  +  b^ 

Trial  divisor,  3  x  a"  =  3  a'^. 

Second  term  of  root,  3  a%  ^3a^  =  b. 

Complete  divisor,       '3a^-\-3ab  +  b^. 

3a%-^'3  ab^  +  b^ 
3  a^ft  +  3  a62  +  J8 

Ex.  1.   Find  the  cube  root  of  27  x«  -  54x4|/8  +  36  xY  -  8  y^- 
The  process  may  be  carried  out  as  follows  : 

Given  Expression  Cube  Root 

27a;6  -  b4xY  +  36xY  -  81/  |  Sx^  -  2/ 
27a;6 


First  term  of  root,  'V^27  a;«  =  3x^. 
(3a;2)8  = 


Trial  divisor,    3(3x2)»  =  27a;*. 
Second  term  of  root,  —54x*y^  -^^x*  =  —  2y\ 
Complete  div.,  3(3  a;2)2  +  3(3  x^)'—  2  2/3)+(-.2  y^)^. 
(27  y*  -  ISrcV  -|-  4g/«)(—  2y3)  ■=. 


54a;Y  +  36a;V-8i/'> 
54a:V  +  36a:V-8?/' 


380 


FIRST  COURSE  IN  ALGEBRA 


28.  It  may  be  shown  that,  by  properly  grouping  the  terms,  the 
cube  root  of  any  polynomial  cube  may  be  obtained  by  repeating  the 
steps  of  the  process  for  finding  the  cube  root  of  a^  +  8  a^b  +  3  aO'^  +  6*. 

The  cube  of  a  polynomial  of  three  terms  may  be  arranged  as 
follows : 


(a  4-  6  +  c)»  =  (a  +  b)*  +  3(rt  -f  6)%  +  3(rt  +  b)c^  +  c«, 


=       a» 


=      a* 


+  3a2& 
+  3  (162 
+      b* 

+  3a« 
+  3a6 

+      b^ 


+  3(rt  +  6)% 
+  3(rt  +  6)c2 
+  ^, 

+  3(a  +  /0'* 
+  3(a+6)c 
+  C' 


Similarly  for  four  terms  : 


(a-[-b  +  c+dy  = 


+  3a2 
+  3rt6 
+     b^ 


+  3(«  +  6)2 
+  3(a  +  6)c 

+  c' 


c  +  3(a-\-b-\-cy 
+  3(a  +  6  +  c)d 
+  rf2 


(0 
(2) 

(3) 
[(4) 


In  (3)  and  (4)  the  expressions  at  the  left  of  the  vertical  bars  are 
the  complete  divisors  used  in  the  extraction  of  the  cube  root  at  the 
successive  stages  of  the  process. 

29.  From  §§  27  and  2.S  it  appears  that  we  can  find  the  cube  root 
of  a  polynomial  cube  of  any  number  of  terms  as  follows  : 

Rule  for  finding  tlie  principal  cube  root  of  a  polynomial 
cube. 

Arrange  the  terms  of  the  given  expression  and  all  successive  re- 
mainders according  to  descending  or  ascending  powers  of  some  letter. 

For  the  first  term  of  the  required  root  write  the  cube  root  of  the 
first  term  of  the  arranged  expression^  and  subtract  its  cube  from  the 
given  expression  to  obtain  the  first  remainder. 

To  obtain  the  next  term  of  the  root,  divide  the  first  term  of  the 
arranged  first  remainder  by  a  trial  divisor  which  is  three  times  the 
square  of  the  part  of  the  root  already  found. 

Construct  a  com^plete  fUvisor  by  adding  to  the  trial  divisor  three 
times  the  product  of  the  part  of  the  root  previously  found  and  the 
term  of  the  root  last  obtained,  and  add  ako  the  square  of  the  term 
of  the  root  last  obtained. 


ETOLUTION  ^81 

Subtract  from  the  first  remainder  the  product  obtained  by  multi- 
plying the  complete  divisor  by  the  term  of  the  root  last  obtained. 

Repeat  these  steps,  with  successive  remainders,  until  zero  is  ob- 
tained as  a  remainder. 

Ex.  2.  Find  the  cube  root  of  8  a:«  -  36  x^  +  66  a:*  -  63  x^  +  33  a:^  _  9  x  +  1. 
Arranged  according  to  descending  powers  of  x,  we  have : 

Given  Expression  Cube  Root 

8ar«-36a;5+66a;4-63x3+33a;2-9x+l  1 2x2-3a;+l 

2^8  =  Sa,6 


First  term  of  root,  ^^x^=^x^. 


(2x2) 


Trial  divisor,  3(2a;2)2  =  nx^. 
Second  term  of  root,  —  SCxf'  -^  12x«  =  —  3x 
Completediv.,3(2x2)-^+3('2a;2)(— 3j-)+(— 3a;)2 
(12j^  —  18x3  _|_  9j;2)(—  3x)  = 


-36x6 
-36x5+54x<-27x8 


Trial  divisor,  3(2x2  _  Zxy  =  I2x*  —  36a^  -f  27x2. 
Third  term  of  root,       12x*  -^  12x4  =.  i 
Complete  div.,  3(2xS  —  3x)2  -|-  3(2x2  _  3^.)  i  _|_  (i)2. 
(12x*— 36x3  +  33x2  — 9x  +  1)1  = 


+12a;*-36x« 

+1 2x-*-36x3+33x2_9x+l 


Exercise  XVIII.    5 
Find  the  cube  roots  of  the  following  expressions  : 

1.  27a;«  +  b^x^ -^  36  ic  +  8. 

2.  8  a;*  +  84  x^y  +  294  a;/  +  343  y^. 

3.  512  a»  -  1344  a^^  +  1 1 76  ab^  -  343  ^>». 

4.  8  a«  -  36  a^b  +  66  a*^**  -  63  a^b^  +  33  a^'b^  -9ab'  +  b\ 

5.  125a^— 225«'+150«Hl35a^-180«*+  33a»  +  546«'-36a+8. 

6.  343aH441a^6  +  777«*^'+531«'/^''  +  444«'^*+144«6^^-64  6^ 

7.  729 «•  +  972 a^?/  +  918 «*/  +  496 a^y  +  204 xhf  +  48 xy^-\-S y\ 

8.  a'' -  'da^^ -  3a«  +  6a'  +  8««  +  3a^ - 3a*-  7a'- 6^2 - 3a- 1. 

27  ^  6  ^  3  ^  8  ^  4  ^  32  ^  64 

Square  Roots  of  Arithmetic  Numbers 

30.  Any  arithmetic  number  may  be  written  in  the  form  of  a 
polynomial  whose  different  terms  contain  different  powers  of  10, 
and  hence  the  method  for  extracting  the  square  root  of  a  polynomial 
may  be  applied  to  arithmetic  numbers. 


382 
E.g. 
Or, 


FIRST  COURSE   IN  ALGEBRA 

625  =  6  •  100  +  2-10  +  5 
=  6  •  102  ^  2-10  +  5 
=  4  •  102  +  20  •  10  +  25. 


Ill  this  last  form,  in  which  the  coefficients  of  the  first  and  last  terms  are 
squares,  the  square  root  of  625,  considered  as  a  polynomial,  may  be  found 
by  the  algebraic  process. 

Given  Expression.  Square  Root. 

2-10  +  5 


First  term  of  root,  ^4-102  =2-10. 

4  •  102  ^  20  • 

10  +  25 

(2  •  10)2  _ 

4  •  102 

Trial  divisor,  2(2  •  10)  =  4  •  10 

+  20 

10  +  25 

Second  term  of  root,  20  •  10  +  4  •  10  =  5. 

Complete  divisor,            4  •  10  +  5 

(4  •  10  +  5)5  = . 

20 

10  +  25 

The  square  root  of  625  is  thus  found  to  be  2  •  10  +  5,  that  is  25. 
31.  Consider  the  following  relations  between  powers  and  roots 


Since 


12=1, 

102  _  100^ 
1002  _  10000, 


10002  ^  1000000, 
etc. 


we  have  -y/l  =  1. 

4 // any  number  between \ /a  number  between \ 

'V  1  and  100  J~\         1  and  10  / 


Vioo 


10. 


W/ any  number  between  \ /a  number  between \ 

'V       100  and  10000       /       V        10  and  100        / 

Vioooo 


=  100. 


t/Zany  number  between \  _  /a  number  between \ 
'  V    10000  and  1000000    ^^       100  and  1000      / 


/y/lOOOOOi.) 


=  1000. 

etc. 


From  the  relations  above  it  may  be  seen  that  the  square  of  an 
integral  number  having  a  specified  number  of  figures  may  be  ex- 
pressed as  an  integral  number  having  either  twice  as  many  figures 
as  the  given  number,  or  as  one  having  a  number  of  figures  one  less 
than  twice  as  many. 

Accordingly,  the  number  of  figures  in  the  integral  part  of  the 
square  root  of  an  integral  number  may  be  found  by  separating  the 
figu/res  of  the  given  integral  number  into  groups  of  two  figures  each, 
beginning  at  units  place. 

The  number  of  figures  in  the  square  root  will  be  equal  to  the  num- 


EVOLUTION  383 

her  of  groups  thus  obtained^  provided'  that  any  single  figure  which 
remains  on  the  left  is  counted  as  a  complete  group. 


E.  g.  The  number  of  ficjurea  in  the  square  root  of  974169  is  three;  in 
the  square  root  of  5  480281  is  four. 

Ex.  1.   Find  the  square  root  of  3249. 

Since,  beginning  with  the  units'  figure  9,  we  can  separate  the  figures  into 
two  groups  of  two  figures  each,  32  49,  it  appears  that  the  integral  part  of  the 
required  square  root  must  be  a  number  of  two  figures. 

Any  given  number  expressed  in  the  common  system  of  arithmetic  nota- 
tion having  two  or  more  figures  may  be  regarded  as  being  composed  of  a 
certain  number,  ^,  of  tens,  increased  by  some  number,  w,  of  units.  Hence  we 
may  represent  any  number  by  <  •  10  +  u. 

Accordingly,  \i  t  -  \0  -\- u  represents  the  number  which  is  the  required 
square  root  of  the  given  number,  3249,  we  may  represent  the  square  of  the 
square  root,  that  is  3248,  by 

{t '  10  +  w)2  =  (^  .  10)2  +  2  (MO)  w  +  w2 
=  <2  (100)   +  2  «w  •  10  +  u^ 

Since,  depending  upon  the  value  of  t,  ^^(loo)  represents  a  number  having 
three  or  four  figures,  it  appears  that  for  this  example  t  must  have  such  a 
value  that  its  square  f^  shall  not  be  greater  than  32. 

The  square  next  4ess  than  32  is  25,  hence  we  assume  that  f^  represents 
25,  or  that  t  =  b. 


c 

iven  Number 
3249     1 
25  00 

Square  Root 

0.10)2;               (5 

102   =  (5  +  )  .  10 
10)2  ^ 

5.  10  +  7 

2(«.10);         2(5 
749  -f  100  =  7  + 

2  (<  .  10)  +  w  ; 
"2(i.  10)-+-w]i*; 

10)    =  100. 

7=      7 
107 
107  X  7  = 

7  49 
7  49 

Observe  that  the  first  remainder  749  contains  the  two  terms,  2  (t  .  10)  u 
and  u^,  combined  into  one  arithmetic  sum.  Accordingly  we  cannot,  as  in 
the  algebraic  process,  obtain  at  this  step  the  exact  value  of  the  next  term 
of  the  root  represented  by  w,  by  dividing  749  by  100. 


384  FIRST  COURSE  IN  ALGEBRA 

It  should  be  understood  that,  in  the  arithmetic  process  above,  the  quotient 
7  +  suggests  but  does  not  definitely  determine  the  value  of  the  root  figure 
sought.  We  may  cissume  7  as  the  root  figure  desired,  and  determine  the 
accuracy  of  our  assumption  by  the  next  remainder. 

Since  in  this  example  the  second  remainder  is  zero,  we  find  that  7  is  the 
second  figure  of  the  re([uired  square  root,  5-10+7,  which  is  57. 

If  in  the  first  step  of  this  example  the  square  less  than  32  had  been 
assumed  to  be  16  instead  of  25,  the  first  root  figure  would  have  been  4 
representing  40,  instead  of  5  representing  50. 

In  this  case  the  integral  part  of  the  (quotient  resulting  from  the  division 
of  the  corresponding  remainder  1649  by  the  trial  divisor  80  would  have 
been  a  number  of  two  figures,  20  +,  instead  of  a  number  equal  to  or  less 
than  9. 

This  would  have  indicated  that  the  square  assumed,  16,  was  too  small. 


32.  Since  there  are  twice  as  many  figures  at  the  right  of  the 
decimal  point  in  the  square  of  a  decimal  fraction  as  there  are  in  a 
given  decimal  fraction,  it  follows  that  the  number  of  figures  at  the 
right  of  the  decimal  point  in  the  square  root  of  a  given  decimal  frac- 
tion may  be  found  by  separating  the  figures  of  the  given  decimal 
fraction  into  groups  of  two  figures  each  from  left  to  right,  beginning 
at  the  decimal  point. 

33.  When  finding  the  square  root  of  an  arithmetic  number  it  is 
often  convenient  to  refer  to  the  formula 


Vt""  +2tu-\-u'=  Vf'  -^  (:2t  ■i-u)u  =  t  +  u. 

At  any  stage  of  the  process,  t  represents  the  part  of  the  root 
abready  found  considered  as  representing  tens,  and  u  represents  the 
next  figure  of  the  root,  which  accordingly  may  be  considered  as 
representing  units  with  reference  to  the  part  of  the  root  already 
found. 

Thus,  it  may  be  seen  that  we  may  use  2  t  and  2t  i-  uto  suggest 
the  different  trial  and  complete  divisors  respectively.  It  should  be 
observed  that  in  each  case  the  number  represented  by  t  is  to  be 
multiplied  by  10. 


EVOLUTION 


385 


Ex.  2.    Find  the  square  root  of  3642.1225. 


V«2.     ^2(5   =6. 
t^;           62  = 

.3642.1225 
36 

2  t ;  2(60;  =  120. 
42  ~  120  is  not  au  integer. 
Hence  root  figure  is  0. 

42 

21;  2(600)  =  1200. 
4212  -f  1200  =  3  + 

42  12 

u]              3=        3. 

2t  +  u;            1203. 
(2«  +  w)w;        1203x3  = 

36  09 

2 «;  2(6030)=  12060. 

6  0325 

60325  -f  12060  =  5  + 

m;              5=          5 

2t-\-U;             12065. 

(2«  +  7z)w;        12065  x  5  = 

6  0325 

Given  Number    Square  Root 
I  60.35 


34.   The  examples  of  §§  31,  33  illustrate  the  following  rule  : 

Separate  tlie  figures  of  the  given  number  into  groups  oftwofigwes 
each^  beginning  at  units'  place. 

Find  tlie  greatest  square  which  is  not  greater  than  the  number 
represented  by  the  figures  in  tlie  first  group  at  the  left^  and  write  its 
square  7'oot  as  the  first  figure  of  the  required  roof. 

Subtract  the  square  of  tlie  root  figure  thus  found  from  the  first 
group  at  the  left,  and  to  the  remainder  annex  the  next  group  of  figures 
for  a  ^^  first  remainder  '\ 

Divide  this  "  remainder  "  by  a  trial  divisor  which  is  obtained  by 
taking  twice  the  part  of  the  root  already  found,  considered  as  repre- 
senting tens,  and  write  tlie  integral  part  of  the  quotient  as  the  next 
root  figure. 

Add  the  root  figure  last  found  to  the  trial  divisor  to  form  a  cojn- 
plete  divisor,  and  multiply  this  complete  divisor  by  the  figure  of  the 
root  last  foundj  subtracting  the  product  from  the  "  remainder  "  last 
found  to  obtain  a  new  ** remainder  '\ 

Bepeat  these  steps  with  successive  groups  of  figures  either  until  all 
of  the  groups  have  been  used  or  until  as  many  figures  have  been  ob- 
tained for  the  root  as  are  desired. 

25 


386  FIRST  COURSE   IN  ALGEBRA 

Formula    Vtf+WtuTt^  =  V^T~(2tTu)u  =  t  +  u, 

35.  Whenever,  during  the  process  of  extracting  the  square  root, 
the  product  of  the  complete  divisor  multiplied  by  the  figure  of  the 
root  last  found  is  greater  than  the  remainder  last  found,  it  is  neces- 
sary to  choose  a  lower  root  figure  and  construct  a  new  complete 
divisor. 

36.  The  number  of  figures  at  the  right  of  the  decimal  point  in 
the  square  root  of  a  decimal  fraction  is  equal  to  the  number  of 
groups  of  two  figures  each  at  the  right  of  the  decimal  point  of  the 
given  number. 

37.  Since  the  position  of  the  decimal  point  in  the  square  root 
of  a  decimal  firaction  may  be  determined  by  means  of  the  number 
of  groups  of  two  figures  each  at  the  left  of  the  decimal  point  in 
the  given  number  whose  root  is  to  b.e  found,  it  follows  that  we  may 
disregard  the  decimal  point  altogether  when  constructing  the  dif- 
ferent trial  and  complete  divisors. 

38.  The  square  root  of  a  fraction  may  be  obtained  either  by  find- 
ing the  square  roots  of  the  numerator  and  denominator  separately, 
or  by  first  reducing  it  to  a  decimal  fraction  and  then  extracting  the 
square  root. 


written  as 


The  number  3642.1225  of  example  (2)  §  33  might  have  been 
36421225 


10000 


Hence,  ^36421225  =  J^^^  =  ^  =  60.35. 

'  ^  y       10000  100 

Exercise  XVIII.     6 
Find  the  square  roots  of  the  following  numbers  : 

1.  2209.  7.  64.1601.  13.  49.61511844. 

2.  6241.  8.  4.008004.  14.  .3603841024. 

3.  26244.  9.  4096.256004.  To  four  decimal  places 

4.  64009.  10.  4.141225.  15.   1.00001. 

5.  643204.  11.  .00009409.  16.  10000.00001. 

6.  6625.96.  12.   1.00020001.  17.  59. 


EVOLUTION  387 

Cube  Roots  of  Arithmetic  Numbers 

39.  Consider  the  following  relations  between  powers  and  roots  : 

Since 

1^  =  1,  we  have    y^l  =  1. 

i//  any  number  between  \  __  /a  number  between \ 
'V  1  and  1000  /        V  1  and  10  / 

103  _  1000,  /y/lOOO  =  10. 

^' /  any  number  between  \  -—  (^  number  between  \ 
'V       1000  and  1000000     /        V        10  and  100        / 

1003=1000000,  /y^lOOOOOO  =100. 

.  »//  any  number  between  \  -—  /  a  immber  between\ 

'  V 1000000  and  1000000000/       V      100  and  1000      / 

10008=1000000000,  ^1000000000  =1000. 

etc.  etc. 

From  the  relations  above  it  may  be  seen  that  the  number  of 
figures  in  the  integral  part  of  the  cube  root  of  an  integral  number 
may  be  found  by  sepai'at'mg  the  figures  of  the  given  integral  number 
into  groups  of  three  figures  each,  beginning  at  units  place. 

E.  g.  The  number  of  figures  in  the  cube  root  of  638  277  381  is  three  ; 
in  the  cube  root  of  1  728  is  two. 

40.  If  t  represents  the  number  of  tens  and  u  the  number  of  units 
in  terms  of  which  an  arithmetic  number  may  be  regarded  as  being 
expressed,  a  process  for  finding  the  cube  root  of  an  arithmetic 
number  may  be  developed  by  referring  to  the  identity 

^/t""  +  3  ^%  +  3  tu^  +  m'  =  ^/t^  +  (3  ^=^  +  3  ^M  +  u'')u  =  t  +  u. 

Ex.  1.    Find  the  cube  root  of  79507. 

Since,  beginning  with  the  units'  figure  7  to  separate  the  figures  into  groups 
of  three  figures  each  we  obtain  two  groups,  79  507,  we  find  that  the  integral 
part  of  the  required  cube  root  must  be  a  number  of  two  figures. 

If  the  required  cube  root  is  expressed  as  a  number  consisting  of  t  tens 
plus  u  units,  then  its  cube,  that  is,  79507,  must  be  represented  by 

(«  •  10  +  uy  =  (t  '  10)8  +  3  (^  .  10)%  +  3  (/  •  10)m2  +  u» 
=  <8  •  1000  +  3  tH  •  100  +  3  tu^  -10  +  ^8 
=  «8  .  1000  +  [3^2  .  100  +  3«w  •  10  +  ii^]u. 

From  the  term  t^  •  1000  it  appears  that  t^  must  be  a  cube  of  which  the 


388  FIRST  COURSE  IN  ALGEBRA 

value  is  not  greater  than  the  number  represented  by  the  figures  in  the  first 
group  at  the  left  which  is  79. 

The  integral  cube  next  less  than  79  is  64,  hence  we  assume  that  t^  repre- 
sents 64,  or  that  <  is  4. 


•  10 

Given  Number 

79507     [ 
64  000 

Square  Root 

^t»  •  1000  ;     ^^79  •  1000  =  (4  +) 

4-10  +  3 

(M0)8;                  (4    10)»  = 

3««-100;             3 .  4* •  100  =  4800 

16  507 

15507  -f  4800  =  3  + 

Ztu.  10 j            3 -4- 3- 10=    360 

w^;                           3«=        9 

3e2.100  +  3^Jt.  10+ u^;        5169 

(3/2.i0O  +  3<u.l0  +  w2)w;  5169 

x3  = 

15  507 

It  should  be  understootl  that  the  quotient  3  +  obtained  by  dividinjif  the 
remainder  15507  by  the  trial  divisor  4800  suggests  but  does  not  definitely 
determine  the  value  of  the  next  root  figure,  3. 

Whenever  the  root  figure  thus  found  leads  us  to  construct  a  complete 
divisor  which  when  multiplied  by  this  root  figure  produces  a  product  which 
is  greater  than  the  remainder  from  which  it  i^to  be  subtracted,  it  i«  neces- 
sary to  try  the  next  less  integer  and  construct  a  new  complete  divisor. 

41.  Since  there  are  three  times  as  many  figures  at  the  right  of 
the  decimal  point  in  the  cube  of  a  given  decimal  fraction  as  there 
are  in  the  given  decimal  fraction,  it  follows  that  the  number  of  figures 
at  the  right  of  the  decimal  point  in  the  cube  root  of  a  given  decimal 
fraction  may  be  found  by  sejmrating  thejigu7'es  of  the  given  decimal 
fraction  into  groups  of  three  figures  ecu^h  from  left  to  right,  beginning 
at  the  decimal  pmlnt. 

42,  When  constructing  the  trial  and  complete  divisors  during  the 
process  of  finding  the  cube  root  of  an  arithmetic  number,  it  will  be 
convenient  to  refer  to  the  formula 

\^t^  +  3  ^^w  -f  3  tu^  +  u^  =  \^t^  +  (8  t^+  Stu  +  u^)u  =  t  +  u, 

in  which,  at  any  stage  of  the  process,  t  represents  the  part  of  the 
root  already  found,  considered  as  representing  tens. 

Thus,  the  expressions  3 1^  and  St^  +  3tu  -{-  u^  may  be  used  to 
suggest  the  different  trial  and  complete  divisors  respectively. 

The  number  represented  by  t  should  in  each  case  be  multiplied 
by  10. 


EVOLUTION 


389 


Ex.  2.    Find  the  cube  root  of  12.326391. 


'V^12  =  2  + 
28  = 


Given  Number 
12.   3iS  391 


Cube  Boot 
12.31 


3^2.     3(2  .  10)2   _  1200. 
4326  -f  1200  =  3  + 

3tu  ;       3  (2  .  10)  3=    180 
162;  32^        9 


(3t^-\-Stu-h  w2)  u  ; 


1389 

1389  X  3  = 


4   326 


4    167 


3«2;  3(23-  10)2=158700. 

159  391 

159391  ^  159700  =  1  + 

3  tu  ;  3(23  •  10)  1  =   690 

1*2  ;        12  =     1 

3«24-3«w  +  it2;        159391 

(3  <2  +  3  tu  +  w2)  lA ;      159391  x  1  = 

159  391 

Ex.  3.    Find  the  cube  root  of  204336469. 

When  carrying  out  the  work  we  find  that  after  multiplying  the  first 
complete  divisor  8764  by  the  figure  of  the  root  last  found  8,  and  subtract- 
ing the  product  70112  from  the  first  "remainder"  79336,  a  new  "re- 
mainder" 9224  is  obtained  which  is  greater  than  the  complete  divisor 
8764.  However  the  complete  divisor  constructed  by  using  9  as  a  root 
figure  instead  of  8  is  not  contained  in  the  corresponding  "remainder" 
9  times. 

The  student  should  carry  out  the  process. 

43.   The  examples  of  §§  40,  42  illustrate  the  following  rule  : 

Separate  the  figures  of  the  given  number  into  groups  of  three 
figy/res  each^  beginning  at  units^  place. 

Find  the  greatest  cube  which  is  not  greater  than  the  number  repre- 
sented by  the  figures  in  the  first  group  at  the  left^  and  write  its  cube 
root  as  the  fii'st  fifjure  of  the  required  root. 

Subtract  the  cube  of  the  root  figure  thus  found  from  the  first  group 
at  the  left,  and  to  the  remainder  annex  the  next  group  of  figui-es  to 
obtain  a  ^^  first  remainder" 

Divide  this  "  remainder ''  by  a  trial  divisor  which  is  obtained  by 
taking  three  times  the  squai-e  of  the  part  of  the  root  already  found, 
considered  as  representing  tens,  and  write  the  integral  part  of  the 
quotient  as  the  next  root  figure. 


390 


FIRST  COURSE  IN  ALGEBRA 


Constrtict  a  complete  dlHsor  hy  adding  to  the  trial  divisor  three 
times  the  product  of  the  jxirt  of  the  root  preinously  founds  conmdered. 
as  representing  tens,  multiplied  hy  the  root  figure  last  founds  and  add 
also  the  square  of  the  root  figure  last  found. 

Multiply  this  complete  divisor  hy  the  root  figure  last  found  and 
subtract  the  product  from  tha  remainder  last  found  to  ohtain  a 
new  **r€inain<1er  ". 

Repeat  these  steps  with  Ruccesmtfe  groups  of  figures  either  until  all 
of  the  groups  of  figures  have  heen  used  or  until  as  many  figures  of  the 
root  have  been  obtained  as  are  deMred. 

Formula  \/t^ +SthA-\-S tu^  +  m«  =  v^«»  +  (St^  +  Stu  +  u^)u  =  t  +  u. 

44.  Whenever  the  product  obtained  by  multiplying  the  complete 
divisor  by  the  figure  of  the  root  last  found  is  greater  than  the  re- 
mainder from  which  it  is  to  be  subtracted,  it  is  necessary  to  choose 
a  less  number  for  the  last  root  figure  and  construct  a  new  complete 
divisor. 

Exercise  XVIII.     7 

Find  the  cube  root  of  each  of  the  following  numbers  : 


1. 

32768. 

7. 

28652616. 

To  four  decimal  places : 

2. 

42875. 

8. 

94818816. 

13.  10. 

3. 

68921. 

9. 

569722789. 

14.  34903.588968101. 

4. 

21952. 

10. 

967361669. 

15.  4. 

5. 

140608. 

11. 

448399762.264. 

16.  f 

6. 

1061208. 

12. 

1.003003001. 

17.  h 

THEORY  OF  EXPONENTS  391 


CHAPTER  XIX 

THEORY  OF  EXPONENTS 

Extension  of  the  Meaning  of  Exponent 

1.  In  a  previous  chapter  we  have  defined  an  exponent  as  a  posi- 
tive whole  number,  and  it  has  been  proved  that,  when  m  and  n  are 
positive  integers,  operations  with  numbers  affected  by  exponents 
are  governed  by  the  following  Index  JLaws : 


I. 

Distribution  Formulas 

Bases  Equal 

1  ^  a" 

-tn 

m  >  n, 

7i>m. 

II. 

Association  Formula 

in*")*' 

=   ff»l.U 

= 

{a'y\ 

III. 

Distribution  Formulas    ^  {a 

X  by" 

=  a*"ft*". 

Exponents  Equal 

l^a 

-^  b)"' 

=  a'"  -f- 

6*" 

'■, 

2.  These  laws  were  established  for  positive  integral  exponents 
only. 

It  may  be  shown  by  repeated  applications  of  the  formulas  above 
that  these  fundamental  laws  may  be  extended  to  apply  to  three  or 
more  factors.     We  have  the  following  general  formulas  : 

(i.)     «»"  X  «*»  X  aP  X  ««  X  •  •  •  =  rtm+n+i,+3+ 

(ii.)  ((((a»")'*)^)«)'' =  a^^'P'i'' 

(iii.)  {abed )»"  =  a'"  X  6'"  Xc'"  X  rf"*  X 

3.  According  to  definitions  already  given,  no  meaning  can  be 

attached  to  such  an  expression  as  a*,  for  it  is  absurd  to  speak  of 
taking  «  as  a  factor  one-third  of  a  time. 

Similarly,  since  we  cannot  use  ^  as  a  factor  "minus  two  times," 
we  have  excluded  from  our  calculations  such  expressions  as  6"^, 
aj""",  etc. 


392  FIRST  COURSE  IN   ALGEBRA 

4.  When  applying  the  Fundamental  Index  Laws  we  shall  find 
that  in  certain  cases  exponents  are  obtained  which  are  not  positive 
whole  numbers.  Hence,  in  order  that  these  Fundamental  Index 
Laws  may  hold  without  exception  in  all  cases,  it  is  necessary  to 
remove  the  restriction  that  an  exponent  must  be  a  positive  whole 
number.  We  must  accordingly  investigate  the  meanings  which 
must  be  attached  to  numbers  other  than  integers  when  used  as 
exponents. 

5.  It  is  essential  that  all  exponents,  without  exception,  should 
obey  the  same  fundamental  laws;  hence  we  shall  impose  the  re- 
striction that,  no  matter  what  may  be  the  nature  of  the  number  w, 
a"  mast  hai^e  such  a  meaning  that  In  all  cdse^-  it  obeys  all  of  the 
Fundamental  Laws  qf  Algebra^  and  in  particular  the 

Fuiiclamental  Index  Law    a'"  X  a"  =  a™"^. 

It  will  be  found  that  as  a  consequence  the  remaining  laws  of 
indices  wiU  be  obeyed.     (See  §  1 .) 

Interpretation  of  Zero  and  Unity  as  Exponents 

6.  Whenever  in  the  operation  of  the  index  law  a"*  H-  a"  =  a*""", 
the  exponents  m  and  n  are  equal,  we  obtain  zero  as  an  exponent. 

E.g.  a^-^a*  =  rt^-^  \j^j^yp  =  hP-p 

=  «'.  =6^ 

7.  Since  the  fundamental  law  «"*  x  a**  =  a"*  "*■ "  is  to  hold,  what- 
ever may  be  the  nature  of  the  exponents  involved,  we  have,  taking 
m  equal  to  zero, 

a^  X  rtr"  =  «^  +  "  =  «". 
Hence,  dividing  both  members  by  a",  we  have 

«"  X  a"  -^  a"  =  a"  -^  a" 
Therefore,  a'  =  1. 

Accordingly,  when  a  has  any  finite  value  whatever  different  fi-om 
zero,  a°  is  defined  as  meaning  1. 

E.g.  30=1,  50=1,  xo^l,  (a-j-&)o=l,etc. 

8.  If  7W  —  w  =  1,  we  obtain  unity  as  an  exponent  by  applying 
the  index  law,  a'"  -h  a"  =  a'""". 

E.  g.  a'^  -7-  a^  =  a\ 


THEORY  OF  EXPONENTS  393 

9.  Since,  in  the  illustration  above,  the  quotient  is  the  base,  it  may 
be  seen  that  the  use  of  unity  as  an  exponent  does  not  contra- 
dict our  previous  definitions.  Hence  we  define  the  first  power  of  a 
base  to  be  the  base. 

That  is,  «i  =  a. 

A  Neg-ative  Integer  as  an  Exponent 

10.  Negative  integral  exponents  arise  when  we  attempt  to  divide 
a  given  power  of  a  base  by  a  higher  power  of  the  same  base. 

E.  g.  3*  -^-  36  =  3*-6  =  3-2. 

11.  Since  negative  exponents  are  to  be  subject  to  the  Fundamen- 
tal Index  Law  a'"  X  a"  =  «'"+",  we  have,  if  w  =  —  wz, 

or  X  a-'"  =  rt'  =  1. 
Dividing  both  members  by  dP\  «  ?^  0, 

That  is,  we  define  any  base^  «,  with  a  negative  exponent,  —  m,  as 
being  equal  to  the  reciprocal  of  the  base  with  a  positive  exponent  of 
which  the  absolute  value  is  equal  to  that  of  the  given  exponent. 

12.  Prom  the  identity  «~"'  =  —  it  appears  that  a  negative  ex- 
ponent, —  m,  indicates  that  the  reciprocal  1  fa  of  the  base  a  is  to  be 
used  as  a  factor  m  times. 

Ex.l.       2-=i=i-  E..2.5--i,  =  l. 


Ex.3.      «:=  3', 

a*       a^  ■  a* 

Ex.4.     (ir'  =  j^. 

_  1 

13.   A  negative  integral 

power 

of  zero  is  defined  to  be  an  infinite 

number,  for                     0~* 

,_  1 

1 

14.  Since                    a" 

,^  1 

_n  : 

(1) 

394  FIRST  COURSE  IN  ALGEBRA 

it  follows  that 


1 

a-" 

= 

1 
1 

or 

1 

a-" 

= 

a\ 

That  is,  -^  =  a\  (2) 

Hence,  it  follows  from  (1)  and  (2)  that  a  factor  may  he  trans- 
ferred from  the  numerator  to  the  denominator  of  a  fraction^  or  from 
the  denominator  to  the  numerator ^  proiyided  that  the  quality  of  its  ex- 
ponent is  changed  from  -\-  to  —  or  from  —  to  -\-. 

Write  each  of  the  following  expressions,  using  positive  exponents  only  : 

Ex.  7.  ba-^b^c-*d»  =  ^^' 

Mental  Exercise  XIX.     1 
Find  the  numerical  value  of  each  of  the  following  expressions  : 

26.  a)-«. 

27.  {^)-\ 

28.  a)-«. 

29.  (f)-l 

30.  -(-  iy\ 

31.  .2-\ 

32.  .5-\ 


1. 

2-\ 

2. 

3-». 

3. 

5-^ 

4. 

2-^. 

5. 

3-«. 

6. 

-8"' 

2 

7. 

3-^. 

8. 

2-«. 

9. 

-2- 

2 

10. 

-3-». 

11. 

-(- 

•2)- 

12. 

-(- 

•2)- 

13. 

-(- 

•2)- 

14. 

1 

2-' 

15. 

1 

3-^ 

16. 

1 
5-*' 

17. 
1^ 

1 

(~2)-« 

1 

in 

(-3)- 
1 

-(-4)- 


20   —  •  33-  ■^''■ 

^^-  2-  34.  .1-". 

5  35.  .01-'. 

21-  ^'  36.  .2-". 

.  37.  2». 

22.  i--  38.  5". 


3 


23. 


5-.  39.  - 

2 


2-»         ^«-  !■ 


25.  (i) 


-1 


41. 


THEORY   OF   EXPONENTS  395 

42.  3  X  8".  46.  2'-8''.  ,^         2'' 

49. 


43.  3  +  8^  47.   (ly.  ■  3«  +  4« 

44.  5*^  +  9^  1  50.  (a  +  by. 

45.  10*^  -  G*'.  4'^ '  51.  x'-(x-  y)\ 

Write  each  of  the  following  expressions  with  positive  exponents 

52.  x-\                      _ .      6 «  ^^    b-^cct'^ 
._.  _.  ob. 

53.  y-\ 

54.  -a^>-\                 ^     ar'^b  ^^   ar'b-^ 
75.  87.  — 5-3- 

56.  a-«^l 

57.  —x^y~^ 


74. 

6« 

75. 

7fi 

(/^-^ 

y 

77 

rnr^x 

73.  ^-  85. 


88.  ^^. 


0  7ia~ 


58.  m-^n-^.  ,                             ^     _„,  „ 

59.-^-83^-4.  77. rr--  89. 

60.  dx-^y. 

61.  3a-\  78.  ^-  90.  ^ 

62.  2-i/>. 

63.  3-^^-1  79    Jl".  91 

64. 5. -V  ^-'            ^^^'y' 

&5.-7mx-K  a-'b-'  „^    (m  -  nY 

66.  ^a-%-\  c                                 (x  —  y)'' 

67.  4.x-'yz-\  _^                               3  (g  +  6)-^ 

68.  6-V^^-l  ;2-«   *                             4(a^-7/)-2* 

69.  2c*-«^>-V^  ^^    _^  ^,    6  (c  +  ^)-^ 

70.  6-'a-'b-'.  ^^'        y-''  ^      l(a-by'' 

1  ^y-i  95.  —  (£c  —  ?/)~^ 

a-2  ^'*-  ;s-i«^;  96.  ^-^  +  ^"^ 

79    ?_  ftA    ^  ^^'  "^^  "^  '^~'' 

r»*  ^^-   c^r**  98.  m-'-n-^ 

4^  ^    ^,,-4  99.  x'+2-^x-'. 


c-^  •  ;5w-=^  100.  a-''+2a-'b-^-\-b-\ 


116. 

ace 

b-'d-' 

117. 

1 

a-%-'c 

118. 

1 

119. 

1 

X  +  IJ 

120. 

(a  +  bXx-^y) 

{a-b){x  +  y) 

1Q1 

n{a  +  b)-^ 

396  FIRST  COURSE  IN  ALGEBRA 

In  each  of  the  following  expressions  transfer  the  factors  from  the 
denominator  to  the  numerator: 

101.^-  108.  --3.  115.^ 

*  or  w 

102.  - 
103. 
104. 

105.  V 
106. 

A  Positive  Fraction  as  an  Exponent 

15.  Positive  fractions  as  exponents  arise  from  an  attempt  to  find 

the  rth  root  of  a^  when  r  and  p  are  integers  and  p  is  not  an  exact 

multiple  of  r. 

p 

Since  it  is  absurd  to  speak  of  taking  5  as  a  factor  two-thirds  of  a  time,  it 
becomes  necessary  to  find  a  meaning  for  an  exponent  which  is  a  fraction. 

16.  We  may  obtain  an  interpretation  for  a  positive  fractional 
exponent  n/din  which  the  numerator  n  and  the  denominator  d  repre- 
sent any  finite  positive  integers  and  d  is  different  from  zero. 

We  may  assume  that  the  denominator  d  is  positive  and  that  the 
numerator  n  has  the  sign  of  the  fraction  which  may  be  either  posi- 
tive or  negative. 

If  a  be  real  and  positive,  then  since  fractional  exponents  are  fo 
obey  the  Fundamental  Index  Law  a"'  X  a"  =  a"*^",  we  must  have 


X 

108. 

3 

f' 

a^' 

2 

109. 

3a^, 

z 

z 

a^ 

1 

6»* 

110. 

~^« 

c 

5 

111. 

dr^ 

a-^b^ 

a-* 

6 

112. 

b  ' 

a^r 

1 

8 

X-' 

113. 

a-'h-' 

1 

a^U' 

~?* 

114. 

^ 

THEORY  OF  EXPONENTS  397 

/ d  factors  in  all . 


(J)''  =  J 


X  a''  X  a'^  X X  a" 


»  ,  »  ,  n 


terms  iu  all 


Hence,  U**/^  =  a»*. 

That  is,  restricting  the  roots  to  principal  values,  the  c?th  power  of 

n 

a^  is  equal  to  the  wth  power  of  a. 

n 

Or,  a^  is  the  principal  c?th  root  of  a". 

This  may  be  expressed  by  means  of  the  radical  notation  a/«". 

n  

Hence,  a''  =  Va\ 

n 

In  order  that  a''  shall,  without  exception,  represent  the  same  value 
as  V^",  it  is  necessary  that  the  restrictions  relating  to  roots  ex- 
pressed by  the  radical  notation  hold  for  roots  expressed  by  means 

of  the  fractional  notation  a^l     (See  Chapter  XVIII.  §  18.) 

n 

In  particular,  we  shall  understand  when  «"  is  a  term  of  an  alge- 

n 

braic  expression  that  a"  =  \a\  unless  the  contrary  is  expressly 
stated. 

If  the  signs  of  operation,  +  and  — ,  are  to  be  applied  without  ex- 
ception to  rational  numbers  and  to  irrational  numbers  which  are 
indicated  by  the  notation  of  fractional  exponents,  it  is  necessary 

n 

t^t  a*^  should  be  understood  as  representing  the  principal  value  of 
the  root  unless  it  is  known  that  a  is  the  nth  power  of  a  number 
which  is  not  positive.  In  this  case  the  root  taken  will  be  a  number 
which  is  not  positive. 

E.  g.  If  X  for  some  reason  is  regarded  as  representing  the  square  of  a 
negative  number,  then  Vx?  represented  by  x^,  must  be  a  negative  number. 

In  particular,  [(-  3)2]  ^  =  -  3. 


898  FIRST  COURSE  IN  ALGEBRA 

We  shall  consider  that  a  root  of  a  mouomial  expressed  by  the 
notation  of  fractional  exponents  which  is  not  a  term  of  an  algebraic 
expression  may,  according  to  circumstances,  be  either  positive  or 
negative. 

That  is,  49^  =  ±  7. 

17.  It  should  be  observed  that  the  denominator  of  a  fractional 
exponent  is  the  index  of  the  root  which  must  be  taken,  and  that  the 
numerator  is  the  power  to  which  the  base  must  be  raised. 

In  particular,  if  w  =  1,  a*^  =  ^/a. 

n 

18.  Interpreting  a''  as  meaning  the  principal  dth  root  of  «",  we 
have  by  Chap.  XVIII.  §  22  (iv.), 

Ex.  1.  sS  =  ^^'  =  4. 

It  is  often  better  to  obtain  first  the  indicated  root  of  the  base  and  then 
the  required  power  of  this  result,  rather  than  first  to  raise  the  base  to  a 
power  and  then  obtain  the  indiciited  root  of  this  result, 

Ex.  2.  216^  =  Cv'216)2  ^  g*  =  36. 

19.  To  agree  with  the  interpretation  for  a  negative  exponent,  we 

n  n 

shall  define  a  '^  to  be  the  reciprocal  of  a'^. 

That  is,  a  d  =  ±-. 

ad 

20.  It  should  be  understood  that  the  expressions  "fractional 
power,"  "negative  power,"  etc.,  refer  to  the  exponent  of  the  power 
and  not  to  the  value  of  the  power  itself. 

E.  g.    The  one-third  power  of  8  is  the  whole  number  2.  \ 

That  is,  8*  =  2. 

Also  the  "minus  second  power"  of  3  is  the  fraction  1/9. 

That  is,  3-2  =  f 

21.  Both  terms  of  a  fractional  exponent  may  he  multiplied  by  or 
divided  by  the  same  number  {except  zero)  without  altering  the  value  of 
a  given  expression. 

n  pn 

That  is,  a**  =  a^*^. 


THEORY  OF  EXPONENTS  ^^9 

n 

Let  X  represent  the  value  of  cC\  the  base  a  being  positive. 

Then  x  =  a^. 

Raising  both  members  to  the  dth  power,  we  have, 

xf^  =  a". 
Raising  both  members  to  the  pth  power,  we  have 

Extracting  the  pdth  root  of  both  members, 

pn 

X  =  a^'^, 

n  ^ 

Hence,  a^  =  a^. 

22.  It  should  be  understood  that  because  the  fractional  exponent 
n/dis  equal  in  value  to  the  fractional  exponent  pn  /  pd,  it  does  not 
follow  as  a  conseciuence  that  the  dth  root  of  the  nth  power  of  a  given 
base  is  equal  to  the  pdth  root  of  the  pnth  power  of  the  same  base. 

The  demonstration  of  §  21  depends  upon  the  principles  that 
like  powers  of  equal  numbers  are  eqiuil,  and  like  roots  of  equal  num- 
bers are  also  equal.  It  does  not  depend  upon  the  principle  that 
both  terms  of  a  fraction  may  be  multiplied  by  or  divided  by  the 
same  number  (except  zero)  without  altering  the  value  of  the 
fraction. 

n  pn 

That  is,  it  does  not  follow  that  a'^  =  a^'^  because  nfd=  pn/pd. 

The  laws  governing  operations  with  and  upon  fractions  which  are 
factors  of  the  terms  of  an  expression  must  be  shown  to  apply  to 
fractions  which  are  used  as  exponents  before  they  can  be  applied  to 
transform  fractional  exponents. 

23.  It  may  be  seen  that  if  the  restriction  be  removed  that  prin- 
cipal roots  only  be  taken,  the  principle  of  §  21  does  not  always 
hold. 

E.  g.  92  has  a  value  which  is  different  from  9^  if  any  other  than  the  prin- 
cipal values  of  roots  be  taken. 

For  9^  =  (9*)^  =  (6561)*  -  ±  81,  while  92  =  4-  81. 

If  only  principal  roots  be  taken  it  may  be  seen  that  9^  has  the  single 
value  +  81,  which  is  equal  to  the  value  of  9^. 


400  FIRST  COURSE  IX  ALGEBRA 

To  find  the  value  of  9^  we  would  commonly  proceed  as  follows: 

9^  =  9^  =  81. 
From  the  illustrations  above  it  may  be  seen  that,  for  principal  values  only 
of  roots,  9^  =  9*. 

That  is,  if  a  fractional  exponent  is  not  in  lowest  terras,  by  taking 
only  the  principal  values  of  roots  we  shall  obtain  the  same  result 
as  by  reducing  the  fractional  exponent  to  lowest  terras  and  then 
proceeding  with  the  transformed  expression. 

24.  Whenever  a  fractional  exponent,  njd,  is  negative,  we  shall 
understand  that  d  is  positive  and  n  is  negative. 

E.  g.    The  expression  9"  ^  will  be  understood  as  meaning 

9-i  =  9^'  =  (9-i)i=:a)^  =  ±i- 

Mental  Exercise  XIX.    2 
Find  the  value  of  each  of  the  following  expressions  : 

1.  9^.  7.  49~i  13.  64"^.  19.  (1^)1 

2.  8^  8.  64"i  14.  12ll  20.  Gly)"*. 

3.  lei  9.  8X1  15.  I69"i  21.  (|1)^. 

4.  4i  10.   lOO'l  16.  125~i  22.  .25"^. 

5.  Sl  11.  125"l  17.  (D*  23.  .343"l 

6.  25i  12.  144"i  18.   (§t)"^  24.  .008"^ 

Express  by  the  radical  notation  the  following  relations  which  are 
expressed  by  the  notation  of  fractional  exponents  : 


25. 

aK 

26. 

b\ 

27. 

A 

28. 

d\ 

29. 

a=-i. 

30. 

y-i. 

31.  A  37.  e.  43.  y  "\ 

3  -                                                          — 

32.  x^.  38.  a^.  44.  z^ . 

33.  /.  39.  6^.  45.  w^. 

a  m 

34.  z^.  40.  c^.  46.  3'. 

m  5                                                2 

35.  a^,  41.  d^.  47.  2''. 

3  6^                                                   n 

36.  h\  42.  z^"".  48.  af +  i. 


THEORY  OF  EXPONENTS  401 


49. 

y  '•"^^ 

61. 

x'^f'z^ 

73. 

Ic^d^. 

85. 

x-T-y^. 

50. 

m  —  1 

62. 

a'n-^c-K 

74. 

11  abi 

86. 

1 

51. 

ahK 

63. 

a~'b''c~'. 

75. 

ab\ 

87. 

(a  +  bf. 

52. 

m^n^. 

64. 

SxK 

76. 

Qib^)^. 

88. 

(c.~d)\ 

53. 

hKX 

65. 

ryyK 

77. 

{ccPf^. 

89. 

(:x  +  y)\ 

54. 

1        1 

66. 

4-1 

78. 

(^¥)'. 

90. 

(z  -  ^)l 

55. 

a'^'x'K 

67. 

7w^. 

79. 

(^V)l 

91. 

(m  —  n)~'^'. 

56. 

ahK 

68. 

1 
—  Ha\ 

80. 

»  +  2 

2  '  . 

92. 

(a'  -  b')i 

57. 

1        _1 

69. 

2c{ 

81. 

1-A 

93. 

(a«  +  />«)3. 

58. 

xUj\ 

70. 

10m^\ 

82. 

2^bK 

94. 

(c^-4)*. 

59. 

h-^y-\ 

71. 

I2ahi 

83. 

3  -^  4  cl 

95. 

(b  +  cf. 

60. 

1      1 

-a    'b   ^. 

72. 

-Ga%i 

84. 

-6 -^11  6?^.  96. 

(a  +  bY  +  \ 

Express  by  the  notation  of  positive  fractional  exponents  the  roots 
which  in  the  following  expressions  are  indicated  by  the  radical 
notation  : 


97. 

V^. 

109. 

4^^. 

121. 

^6-^. 

130. 

-^abh\ 

98. 

^6. 

110. 

-  5^p. 

122. 

^^. 

131. 

Va'^-'b. 

99. 

^c. 

111. 

la\/Fc. 

123. 

^/x--^. 

132. 

2^a\ 

100. 

^x. 

112. 

e^xy. 

124. 

^l^x-^ 

133. 

-  a</lf. 

101. 

^^. 

113. 

Vl  -=-«. 

125. 

5\^ab^c. 

134. 

"^Vm. 

102. 

-^r^. 

114. 

—  Vl-r-C. 

126. 

m'^  m^n^w 

'  135. 

\l\^n. 

103. 
104. 
105. 

V-x^ 

115 
116. 
117. 
118. 

\^l^b\ 
-</l-^d^ 

'^2  -^  c\ 

127. 

128. 

136. 
137. 
138. 

yjaVb. 
-  a^b^/'c. 

x^^yWz. 

106. 

Vl  -f-  m. 

107. 

108. 

Wx. 

2^y. 

119. 
120. 

Va-K 

-  Vx-\ 

129. 

m- 

139. 
140. 

402  FIRST  COURSE  IN  ALGEBRA 

Products  of  Powers  and  Quotients  of  Powers. 

25.  The  laws  relating  to  products  of  powers  and  quotients  of 
powers  of  the  same  base  may  be  applied  without  exception  when 
exponents  are  zero,  negative  numbers,  or  fractions. 

This  is  because  exponents  which  are  zero,  negative  numbers,  or 
fractions,  were  so  defined  as  to  obey  the  Fundamental  Index  Law 
«'"  X  a"  =  rt'"''"'',  and  accordingly  the  derived  law  a"*  -i-  c*"  =  a'"~". 

Ex.  I.    Simplify         9?  x  27i 

We  have  9^  x  27^  =  (O^)'  x  (27^2 

=  38  X  32 
=  36 
=  243. 


Ex.  2. 

Simplify 

a*  X  a-8. 

We  have 

a^  X  a-8  =  a^* 
-a»' 

Ex.3. 

Simplify 

h-l  X  hi 

We  have 

Ex.  4. 

Simplify 

8c^-i-2ci 

We  have 

Exercise  XIX.     3 
Simplify  the  following  products  and  quotients,  applying  the 

Index  Laws  :  I  ^^,„  _^  ^^„  ^  ^^,„_„ 


L  4*  X  si 

5.  3  X  O"^. 

9.  rt'r«-^ 

2.  25^  X  125^. 

6.  25"^  X  5-1 

10.  b-%-\ 

3.  64^  X  lei 

7.  32^  -^  4^ 

n.  c«c-«. 

4.  4*  X  2"^ 

8.  8~*  X  2-\ 

12.  ...-i 

THEOKY   OF  EXPONENTS  403 


33. 


15.  ^!/^  xiijK  „,  (-  ^y 


13.  (5^)(4^o). 

14.  ^*.r-'^-^  —a 

16.  Sw~^X2w~K  ^~^^ 

17.  m^m^rn*. 

18.  ifn-'tf. 

19.  (.r"-0(^~OC?^'"). 

21.  .r«-T-^-'. 

22.  2/^  --  y-'\ 

23.  C-*  H-  c-^ 

—  £t  lit     y 

24.  r^-^-ri  ,  , 

25.  4s*-T-2A  39.  ^. 

26.  fi?"!-^(^"l  af/ 

27.  o^+'-^-f-S/.  ^  «  -J. 

40.  aj"a;"'a;  '"^ 


35. 

36. 

15«-* 
-3«-^ 

37. 

-20c-V;:-« 

^dn'z-" 

38. 

-  18m-V 

m+n    w» — n 


28.  §  /^P-^  -T-  /^«-P 

29.  -  12ah^  -^  3 ail  41.  x^x^ 

30.  a^6^  -7-  Jbl  ,       ,  -9  -i 

31.  16  w*?/^  -^  8w%^ 

32.  «^6*c;* -=- a*6V.  43.  (b^-^c^)(l>^-^^c  *). 

26.  Since  exponents  which  are  zero,  negative  numbers,  or  frac- 
tions, have  been  so  defined  as  to  obey  the  Fundamental  Index  Law 
a"*  X  a"  =  «"•+",  and  accordingly  the  derived  law  (.r  ~r  a""  =  a"*"", 
it  remains  for  us  to  show  that  with  the  meanings  thus  obtained 
these  exponents  obey  all  of  the  laws  relating  to  positive  integral 
exponents. 

Unless  the  contrary  is  explicitly  stated  we  shall  understand  the 
word  "  exponent "  to  have  any  of  the  meanings  previously  defined. 

27.  From  the  Index  Law  «'"  X  ^"  =  a""^'\  I,  we  may  derive  the 
law  a""  -j-  f//*  =  a"*~",  and  from  the  law  (aby  =  «"  i"  we  may  derive 

thelawg)"  =  |;. 


404  FIRST  COURSE  IN  ALGEBRA 

Hence  it  is  necessary  and  sufficient  for  us  to  show  that  the  index 
laws  (a"*y  =  a*"",  II,  and  (abf  =  oTW,  III,  apply  for  all  commen- 
surable exponents. 

Powers  of  Powers. 

28.*  Proof  that  the  index  law  («"*)"  =  '''""*  applies  for  all  com- 
mensurable exponents. 

(i.)  If  n  be  a  positive  integer,  then  for  all  values  of  tw,  we  have, 

n  factors » 

{a^y  =  (r  X  oT  X  (r  X x  «"* 

(ii.)  Let  w  be  a  positive  fraction,  />  r,  in  which  p  and  r  are  posi- 
tive integers. 

Then  (a")"  =  (fff 

p 
Raising  (a")'  to  the  rth  power,  we  have 

=  {a-y. 

Since  jo  is  a  positive  integer  it  follows  that 

Writing  the  principal  rth  roots  of  both  members,  it  foUows  that 

p  mp 

Hence,  substituting  for  p/r  its  value  w,  we  have 

(iii.)   Let  n  be  any  negative  number  denoted  by  —  q. 

Then  (cry  =  (oT)-^ 

._     1 

=  «-"'. 
Hence,  substituting  for  —  q  its  value  w,  we  have, 

(a"')"  =  a"^. 
(iv.)   Let  n  be  zero. 

Then     {aT'y  =  «"""  is  satisfied  if  a  ^  0  and  n  =  0. 
For,       (a^'y  =  «'"»,  means  1  =  1. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


THEORY  OF  EXPONENTS  405 

It  follows  from  the  reasoning  of  (i.),  (ii.),  (iii.)  and  (iv.)  that  for 
all  commensurable  values  of  the  exponents  we  have 

29.  Substituting  1  /  r  for  w,  and  p  for  n,  it  may  be  seen  that 

from  («"•)"  =  «r  it  follows  that  (a'V  =  {a)\ 

That  is,  the  pth  power  of  the  rth  root  of  any  base^  «,  which  is  differ- 
ent from  zero^  is  equal  to  the  rth  root  of  the  pth  power  of  the  base. 

30.  It  should  be  observed  that  if  the  restriction  regarding  prin- 

1  p         p  1 
cipal  roots  is  removed,  the  relation  (a!")  =  (a  Y  does  not  always 

hold. 

E.  g.    (92)2  ^  9^  but  (92)^  =  ±  9. 

Hence,  (92)2  :^  (92)2  except  for  principal  values  of  the  roots. 

Ex.  1.    Simphfy  (16^^. 

We  have  (16^)-8  =  (2)-» 
_  1_ 
~28 
I 

~8* 

Mental  Exercise  XIX.    4 

Find  simplified  expressions  for  the  following  indicated  powers  of 
powers,  applying  the  Index  Law  («'")"  =  a"*"  : 


1.  (c^)-«. 

11.  {k-'f. 

21.  {h^y. 

2.  {d-^y. 

12.  -(m')^. 

22.  (F'»)^. 

3.   {a-')-\ 

13.  in-"^)^. 

23.  {d^'^y. 

4.   (n-«. 

14.  -0^)i 

24.  (?^*0"^. 

5.  ix-y. 

15.  {x^)^. 

25.  -(O- 

6.  (/)-^. 

16.  (^-^)- 

26.  (a;«^)-^ 

7.  {2^y. 

17.  {z-^yK 

27.   (a'^*. 

8.  {w^f. 

18.  -{w^)^. 

28.  -(^')^ 

9.   (^1°)*. 

19.  _(w2-i)-i. 

29.  (c^-^. 

10.  {ii'Y\ 

20.  —  (tw-'^'^)^ 

30.  (^)-". 

406 

FIRST  COURSE   IN 

ALGEBRA 

31.  {h-y. 

51. 

{by 

71. 

{m^^y 

32.  {k-y. 

52. 

{c-y. 

72. 

{n'-y 

33.  -{my\ 

53. 

{drr. 

73. 

(af+*)'. 

34.  {ny\ 

54. 

{x-")-. 

74. 

(/-^)* 

35.  (ic-0-«\ 

55. 

(f-r 

75. 

{z^-^^y 

36.  (a*)"- 

56. 

-  (-)«^ 

76. 

{w"-y 

37.  -{ify. 

57. 

{w*^)^. 

77. 

(m'"-y\ 

38.  (0-». 

58. 

{a'y. 

78. 

{a''+')K 

39.  ((^-^)'. 

59. 

{b''y\ 

79. 

{b'^-^)K 

40.  (A-"')-*. 

60. 

(c— )-^". 

80. 

(^/"-1«. 

41.  (^)l 

42.  (mr. 

61. 

{{..-yy. 

81. 

{k^y^+\ 

62. 

{{fry- 

82. 

/yJ^+2\.7-2^ 

43.  {ny. 

63. 

{(z^yy. 

83. 

{a'^+y-^ 

44.  (a-)"\ 

64. 

(K)-»)- 

84. 

{af+'y+\ 

45.  {y-y 

65. 

{a-^y 

85. 

(/+«y+^ 

46.  (r^)\ 

66. 

{b'^-y 

86. 

/^7n-2\m+8^ 

47.  (?/;""')". 

67. 

{c^^y 

87. 

{w^-^y\ 

48.  -ia'S*' 

68. 

{d^^^K 

88. 

{b'+y-^'. 

49.  (6«^)-l 

69. 

{hr-'- 

89. 

(c'^'-^)^*. 

50.  {aj. 

70. 

{k'y^^. 

90. 

{<F+y"-y"^\ 

Powers  of  Products  and  Powers  of  Quotients. 

31.*  Proof  that  for  all  commensurable  values  of  the  exponents 
we  may  apply  the  Index  Law  {aby  =  a^"". 

(i.)     Let  w  be  a  positive  fraction. 

The  law  has  been  shown  to  apply  for  all  positive  integral  values 
of  the  exponents. 

♦  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


THEORY  OF  EXPONENTS  407 

Substituting  p/r  hr  nwe  may  write 

(aby  =  (aby. 
p 
Writing  the  rth  power  of  (aft)'",  we  have 

[{abn^iaby.^ 

—  ay¥,  since  jo  is  a  positive  integer. 

Furthermore,  {My  =  {ay  {by  =  ayjf. 

Hence,  [Wl'  =  («V)^ 

Writing  the  principal  rth  roots,  we  have 

{aby  =  a'b\ 
Or,  {aby  =  a%",  when  n  is  a  positive  fraction, 

(ii.)   Let  n  have  any  negative  value  represented  by  —  q. 
Then  {aby  =  {aby^ 

_     1 

-  {aby 

__   1 

~  a'^b'^ 
=  a-^b-^. 
Or,  {aby  =  a%\ 

(iii.)  Let  n  have  the  value  zero. 

Then  {aby  =  a^b^^  because  {(ihf  means  1,  and  a^b^  means  1  X  1 
for  bases  a  and  6,  which  are  different  from  zero. 

It  follows  from  the  reasoning  of  (i.),  (ii.),  and  (iii.)  that  for  all 
commensurable  values  of  the  exponent  we  have 

{aby  =  a^^b^ 

Ex.  1.    Simplify  (ah'^y^ 

We  have  (ah-  ^y^  =  a%-* 


Ex.2.    Simplify  (^/)' 

V//   ~r 


We  have 


y-rz 


408  FIRST  COURSE   IN  ALGEBRA 

Mental  Exercise  XIX.    5 

Simplify  each  of  the  following  expressions,  applying  either  the 
index  law  (aby  =  a"^",  or  the  index  law  (a  -H  by  =  a"  -^  b"  : 


1.  (ahy. 

17.  (a-'  -^  6«)2. 

33.  (2c-y. 

2.  (b'c-y. 

18.  (^^«^cV- 

34.  (3t^«)-^ 

3.  (c-'d*)-^ 

19.  (c-^  H-  </«)-2. 

35.  (4  3!)^. 

4.  (d-^h-'^y. 

20.  (a-^-^x'y\ 

36.  -  (9/)^. 

5.  (m'n-'y. 

21.  (6-^-c-y. 

37.  (lez^yK 

6.  (x-yr'' 

22.  (0-^-3,-1)-^. 

38.  (25  O"^- 

7.  -(a^x^K 

23.  {a--'-^b-y\ 

39.  (S^-^ab-yi 

8.  -  (b'y^y. 

24.  (a^  ^  b^y. 

40.  (d-^m-'n-'yK 

9.  -  ((rh*)K 

25.  -  (6*  H-  x^y. 

41.  (a^^^-V)-^. 

10.  {d'w-^y^\ 

26.  (c* -r- 3^^)-^. 

42.  (bh'd-^y. 

11.  a-V*)i 

27.  (^-i-^)-«. 

43.  (a-^^^O-2. 

12.  (6V^)-«. 

28.  (h-T-m-y^. 

44.  (^«c-V^)-i 

13.  («¥)«. 

29.  (a2*-T-ic-*)i 

45.  {d-^x-YT^- 

14.  (6^c^)^«. 

30.  (b-^'^d^)K 

46.  (2a-«^-V0-l 

15.  (c-^r^)-^. 

31.  (2fl')-^. 

47.  (3-^a;«/r'-)~'- 

16.  (ry)-K 

32.  (4^-^)^ 

48.  (4-^a^6-V^)-«. 

32.  It  is  customary  to  regard  a  fractional  power  a^  as  being 

r 

higher  or  lower  than  another  fractional  power  a%  according  as 
^  —  -  is  a  positive  or  a  negative  number. 

OS 

E.  g.  We  shall  consider  that  x^  is  a  higher  power  than  x^  since  f  —  ^  =  ^, 
which  is  a  positive  number. 

But  xi  is  a  lower  power  than  x  since  |  —  1  =  —  |^,  which  is  a  negative 
number. 

Exercise  XIX.    6 

The  following  examples  illustrate  the  application  of  the  principles 
of  earlier  chapters  to  expressions  whose  terms  involve  negative  and 
fractional  exponents. 


THEORY  OF  EXPONENTS  409 

Multiply  : 

1.  a*  +  b^  by  a/^  +  h^-  5.  a'^  —  h^    by  a~^  —  b^- 

2.  ic^  —  y^  by  a;^  —  ?/2.  g.  ^"i  +  ^^  by  b^  +  c"^ 

3.  a^^  —  y^  by  a:^  —  y~^.  7.  a^  +  a*   by  a^    +  33^ 

4.  x'^  +  y"^  by  a^^  —  y^.  8.  a;-*— 3a;-2+  1  by  aj-^  +  2a;-^ 

9.   2 a;^  —  3 a;^  —  4  +  a;"Hy  3 «*  +  a;  —  2 a;i 
10.  a-%-'  +  2  a-^b''  -  3  a%-^  by  a'^/"^  +  3  a'b. 
Divide : 

11.  a;-'  —  3  x-^  +  2  x"'-'  by  a;-^'  —  2  a;"^  +  1. 

12.  a*  —  a*/>^  —  a^b^  +  ^/^  by  a^  —  b^. 

13.  a;  —  y  by  x^  —  ?/. 

14.  a;-«  -  3  a^-V'  +  3  aj-y  -  y'^  by  a;""  -  y-\ 

15.  a  +  ^  by  a^  +  b^- 

16.  a;"^?/  —  5  xy~^  +  4  a?^^"^  by  xhj-^  +  a;V~^  —  2  aj^"^. 

17.  a  — ^>by«*  — ^/i 

18.  2  a'%'  -  4  a"^/!/^  +  2  by  2  a-%^  —  4  a"*^^  +  2  a"^6l 

Exercise  XIX.     7.     Miscellaneous 
Simplify  each  of  the  following  expressions  : 

1.  4^  X  16"^  X  64i 

16^  X  125"^ 
'  25^  X  32^ 

3.  9°  +  9^  —  9"^  +  9^ 

4.  32<*  +  32^  +  32^  +  32^  +  32^  +  32. 

5.  2(0""  -  («"'')■'  -  («"')"'• 
^    2"+^  —  2-2" 

n+8 


2-2 

4(2"-^)"  •  2-^ 
(2"+^)"^^  •  2"" 


410  FIRST  COURSE  IN  ALGEBRA 

8.  2^Gi-9-«■3»-^^ 

10.   (x^Jf. 

■■■  irh)'- 

13    »"-»" 

14.  [(aj')'~-]A. 

15.  (^-./-'^^: 


17. 


3«+l   y.   3—1   X    g-n 


18.    (((««)   c)i;.)'*^(((^   .y)    «y 
X9.   l^Z^,  X     -^^-^ 


ar^-a-^b      ar^  +  </"■ 


'="•    \a-'b-ab-^)\ab-'  )   '   Vl  +  a^'V 

a4-»      1  fr4-c      1  c+fl      1 

22.    [ti*-*^]''-*  X  [w'^^]'^-*  X  [w"-*]*-*. 


\ 

IRRATIONAL  NUMBERS  411 


CHAPTER  XX 

IRRATIONAL  NUMBERS  AND  THE   ARITHMETIC  THEORY 
OF   SURDS 

I.   Irrational  Numbers 

1.  Two  numbers  are  said  to  be  commensurable,  that  is,  to 
have  a  common  measure,  if  both  can  be  divided  without  remainder 
by  the  same  integral  or  fractional  number. 

E.  g.  The  numbers  9  and  12  are  commensurable  because  both  contain 
3  exactly  as  a  divisor. 

Any  two  whole  numbers,  any  two  fractions  whose  numerators  and 
denominators  consist  of  a  limited  or  finite  number  of  figures,  or  any 
whole  number  and  any  fraction,  are  commensurable  numbers. 

E.  g.  The"  two  fractions  \^  and  f  are  commensurable,  since  they  can  be 
expressed  as  integral  multiples  of  a  fraction  having  as  a  denominator  the 
common  denominator  of  the  two  fractions;  that  is,  \^  can  be  expressed  as 
22  times  ^■^,  and  |  as  57  times  ^. 

2.  Two  numbers  which  are  not  commensurable  with  reference 
to  each  other  are  said  to  be  incoraimensurable. 

3.  Since  any  whole  number  or  any  fraction  is  commensurable 
with  respect  to  unity,  whole  numbers  and  fractions  are  commonly 
called  commensurable  numbers.     (Compare  with  Chap.  XVIII.  §  8.) 

4.  Since  any  whole  number  can  be  expressed  as  a  fraction  it 
follows  that  anj/  number  is  commensurable  if  it  can  be  expressed  as 
the  quotient  of  one  whole  number  divided  by  another  whole  number. 

E.  g.  The  number  5  is  commensurable  because  5  can  be  expressed  as  a 
fraction,  such  as  ^. 

The  square  root  of  5  is  an  incommensurable  number  because  it  cannot  be 
expressed  exactly  by  any  fraction. 


412  FIRST  COURSE  IN  ALGEBRA 

5.  In  order  that  the  rth  root  of  a  commensurable  number  c  may 
be  expressed  as  a  commensurable  number  n,  it  is  necessary  and 
sufficient  that  c  be  the  rth  power  of  some  commensurable  number  n. 

For  if  '^c  =  w,  then  c  =  w*". 

E.  g.  Since  9  is  the  square  of  the .  commensurable  number  3,  it  follows 
that  V^  is  a  commensurable  number.  The  number  10  is  not  the  square  of 
a  commensurable  number.  Accordingly,  the  square  root  of  10  is  an  incom- 
mensurable number. 

6.  The  classification  of  numbers  as  being  either  commensurable 
or  incommensurable  may  be  made  to  depend  upon  the  following 
Principles  Relating^  to  Fractions : 

(i.)    The  rth  power  of  an  integer  is  an  integer. 

For,  since  the  operation  of  division  does  not  enter  into  the  process 
of  involution,  a  fraction  cannot  result  from  using  an  integer  any 
number  of  times  as  a  factor. 

(ii.)  The  rth  power  of  a  fraction  in  lowest  terms  is  a  fraction  in 
lowest  terms.     (See  Chap.  XV.  §  40.) 

It  follows  from  the  reasoning  of  (i.)  and  (ii.)  that: 

(a)  The  rth  root  of  a  positive  integral  number  which  is  not  the 
rth  power  of  another  integer  cannot  be  expressed  eitJier  as  an  integer 
or  a  fraction^  that  is,  as  a  commensurable  number. 

E.  g.  Since  2  is  not  the  square  of  any  whole  number,  it  follows  that 
-v/a  cannot  be  expressed  either  as  a  whole  number  or  as  a  fraction. 

(b)  The  rth  root  of  a  positive  fraction  in  lowest  terms  of  which  the 
numerator  or  denominator,  or  both,  are  not  rth  powers  of  positive 
integers,  cannot  be  expressed  either  as  an  integer  or  a  fraction,  that 
is,  as  a  commensurable  number. 

E.  g.  Since  2  and  5  are  prime  to  each  other,  and  neither  is  the  square 
of  a  whole  number,  it  follows  thaty^f  cannot  be  expressed  as  a  commen- 
surable number. 

7.  It  is  not  possible  to  express  the  value  of  the  rth  root  of  a  num- 
ber which  is  not  an  rth  power  exactly  as  an  integer  or  as  a  fraction 
in  lowest  terms.  Still  it  may  be  shown  that  it  is  possible  to  find 
as  approximate  values  two  fractions,  one  greater  than  and  the 
other  less  than  the  true  value  of  the  indicated  root,  which  shall 


IRRATIONAL  NUMBERS  413 

differ  from  each  other,  and  consequently  differ  from  the  true  value 
of  the  required  root,  by  as  small  a  value  as  we  please. 

(The  following  illustration  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

8.  Different  methods  may  be  applied  to  obtain  approximate 
values  for  a  particular  incommensurable  number,  such  as  V2. 

Since  2  is  not  the  square  of  any  integer,  its  square  root  cannot 
be  expressed  exactly  either  as  an  integer  or  as  a  fraction.  (See  (i.) 
and  (ii.),  §  G.) 

Since  2  is  greater  than  1^  and  less  than  2^  it  follows  that  V2 
must  be  greater  than  1  and  less  than  2. 

That  is,  1=^  =  1<  2  <  4  =  2^. 

Hence,  V2  lies  between  1  and  2. 

Or,  1<  V2  <  2. 

The  following  are  the  only  fractions  having  10  for  a  common 
denominator,  the  values  of  which  lie  between  1  and  2 : 

\h    \h    \h    \h    U^    i§'    \h    if'    H- 
Writing  the  squares  of  these  fractions,  we  obtain, 

m^  m^  m^  m^  m^  u^^  nh  uh  m- 
We  find  that  ({^y  =  m  <^<m  =  (nr- 

Hence,  since  2  lies  between  (1.4)'^  and  (1.5)^,  it  follows  that 

1.4  <  V2  <  1.5. 
The  following  are  the  only  fractions  having  100  for  a  common 
denominator,  the  values  of  which  lie  between  {^  and  {^  : 

Hh  uh  Hh  m^  Hh  uh  Uh  \n^  m- 

We  find  that  the  square  of  j^^  is  less  than,  and  the  square  of  i^§ 
is  greater  than  2. 

That  is, .    (mr  =  nm  <  2  <  nhu  =  (mr- 

Hence,  since  2  lies  between  (1.41)^  and  (1.42)^,  it  follows  that 
1.41  <  a/2  <  1.42. 

Continuing  the  process  indefinitely,  it  is  possible  to  find  two 
fi-actions,  one  greater  than  and  the  other  less  than  the  true  value  of 
a/2,  which  differ  from  each  other,  and  accordingly  from  the  true 
value  of  a/2  by  a  value  less  than  any  assignable  number,  however 
small. 


414  FIRST  COURSE  IN  ALGEBRA 

Consider  the  following  relations  : 


1^ 

=  1 

<2<4 

=  2^ 

1.4^ 

=  1.96 

<  2  <  2.25 

=  1.5« 

1.41^ 

=  1.9881 

<  2  <  2.0164 

=  1.42^ 

1.414'' 

=  1.999396 

<  2  <  2.002225 

=  1.415=' 

1.4142^ 

=  1.99996164 

<'2<  2.00024449 

=  1.4143^ 

1.41421^ 

'  =  1.9999899241  <  2  <  2.0000182084 

=  1.41422^ 

etc. 

etc. 

Since  the  true  value  of  V2  lies  between  any  two  corresponding 
numbers  of  which  the  squares  are  indicated  in  the  first  and  last 
columns  above,  it  follows  that  the  difference  between  the  true 
value  of  V2  and  either  of  the  numbers  must  be  less  than  the 
difference  between  the  two  numbers  themselves. 

Any  number  of  which  the  square  is  indicated  in  either  column 
may  be  taken  as  an  approximation  to  the  true  value  of  V2. 

The  numbers  of  which  the  squares  are  indicated  in  the  first 
column  are  all  less  than,  and  those  of  which  the  squares  are  in- 
indicated  in  the  last  column  are  all  greater  than,  the  true  value 
of  V2. 

Hence,  by  extending  the  process  indefinitely,  as  close  an  approxi- 
mation to  V2  as  may  be  required  can  be  obtained. 

9.  If  we  represent  the  different  numbers,  1,  1.4,  1.41,  etc.,  and  2, 
1.5,  1.42,  etc.,  which  are  approximate  values  of  V2,  by  distances 
firom  a  fixed  point  0  measured  along  a  straight  line,  we  may  sug- 
gest as  in  the  accompanying  figure  that  the  successive  pairs  of 
values,  1  and  2,  1.4  and  1.5,  1.41_  and  1.42,  etc.,  approach  nearer 
and  nearer  to  the  true  value  of  V2. 

1.41  [11.42 

1.4iKl.5 


— ^0 h \r2 ^2— 

10.  By  the  reasoning  employed  in  §§  8,  9,  any  indicated  root  of 
an  incommensurable  number  can  be  obtained. 

It  should  be  understood  that,  although  it  may  not  be  possible  to 
express  the  true  value  of  a  specified  root  exactly  in  terms  of  either 
an  integer  or  a  fi-action,  yet  this  value  is  definite. 


SURDS  415 

11.  In  order  that  there  may  be  no  exception  to  operations 
involving  indicated  roots,  we  shall  assume  that,  even  when  the 
radicand  c  is  not  the  rth  power  of  a  commensurable  number,  the 
following  identity  is  true  : 

E.g.  (V3)^  =  3.  (VA)^  =  t\- 

12.  Two  irrational  numbers  are  said  to  be  equal  or  un- 
equal according  as  their  approximate  values  are  equal  or  unequal. 

13.  By  means  of  the  Principles  of  Variables  and  Limits  it  may 
be  shown  that  the  Fundamental  Laws  of  Algebra  apply  for  incom- 
mensurable as  well  as  for  commensurable  numbers. 

11.   Arithmetic  Theory  of  Surds 

14.  A  rational  number  has  been  defined  as  a  number  which 
can  be  expressed  as  the  quotient  obtained  by  dividing  one  whole 
number  by  another.     (See  Chapter  XVIII,  §  8.) 

E.  g.  The  number  4  which  can  be  expressed  as  a  fraction  such  as  |  is  a 
rational  number. 

Any  fraction  such  as  f  is  a  rational  number. 

A  number  which  is  not  rational  is  said  to  be  irrational. 

E.  g.  Since  ^/2  cannot  be  expressed  exactly  as  a  fraction,  it  follows  that 
'y/2  is  an  irrational  number. 

15.  An  indicated  root  of  a  number  or  expression  is  frequently 
spoken  of  as  a  radical. 


E.  g.   /y/S,  Vis,  V5»  V^  +  y- 

16.  A  radical  expression  is  an  expression  containing  one  or 
more  radicals. 

E.  g.   3  V^^,  V^,  y^  +  1,  Va  -  ^A^ 

17.  A  surd  is  an  irrational  or  incommensurable  root  of  a  rational 
or  commensurable  number. 

E.  g.  ^^%  ''^/It,  a/A  ^^^  ^^^^^  numbers,  for  in  each  case  the  radicand  is 
a  rational  number  which  cannot  be  obtained  by  raising  another  rational 
numl)er  to  a  power  the  exponent  of  which  is  equal  to  the  index  of  the 
indicated  root. 


416  FIRST   COURSE   IN  ALGEBRA 

Observe  that  >y/9  is  not  a  surd,  since  -y^Q  =  3. 

Expressions  such  as  V -v/S  —  ^%  VV*^  +  1,  are  not  surds  in  the  sense 
of  the  definition,  since  the  radicands  y'S  —  ^^2  and  -y/S  +  I  are  incom- 
mensurable numbers. 

18.  Although,  from  the  point  of  view  of  algebra,  Vw  is  an  irra- 
tional function,  yet  it  may  or  may  not  be  arithmetically  irrational 
according  to  the  particular  values  assigned  to  n. 

E.  g.  If  n  =  4,  9,  16, ,  -y/m  is  not  a  surd  ; 

while  if  n  =  2,  3,  5,  7,  •  •  •  •  ,  ^Jn  is  a  surd. 

19.  According  as  the  index  of  the  indicated  root  is  2,  3,  4, 

5, ,  w,  surds  are  classified  as  being  of  the  second  order  or 

qnad/ratic,  as  VS ;  of  the  third  order  or  cubic,  as  V^o  ;  of  the  fourth 
order  or  biquadratic^  as  VlO;  of  the  fifth  order  or  quintiCy  as 
-^ ;  of  the  wth  order  or  n-tic,  as  \/x ;  etc. 

20.  A  single  surd  number  or  any  rational  multiple  of  a  single 
surd  number  is  called  a  simple  monomial  surd  number. 

E.  g.    v^j  3\/2i  2  V^- 

21.  The  sum  of  two  simple  monomial  surd  numbers,  or  of  a 
simple  monomial  surd  number  and  a  rational  number,  is  called  a 
simple  binomial  surd  number. 

E.  g.  V^  +  V^     V^^  +  1- 

The  expression  ^x  +  ?/  is  not  a  simple  binomial  surd  number,  but  is  a 
simple  monomial  surd  number  of  which  the  radicand  is  a  binomial. 

22.  The  principles  established  in  the  chapter  on  Evolution  apply 
for  irrational  as  well  as  for  rational  roots. 

We  shall,  in  the  present  chapter,  consider  such  radicands  only  as 
are  positive,  and  shall  use  only  the  principal  values  of  the  indicated 
roots. 

Reduction  of  Surds  to  Simplest  Form 

23.  A  surd  is  said  to  be  in  simplest  form  when  the  expression 
under  the  radical  sign  is  integral,  and  is  not  a  power  of  a  rational 
expression  the  exponent  of  which  is  equal  to  the  index  or  which 


SUEDS  417 

contains  as  a  factor  any  factor  of  the  index  of  the  indicated  root ; 
and  when  there  appears  under  the  radical  sign  no  factor  raised 
to  a  power  of  which  the  exponent  is  equal  to  or  greater  than 
the  index  of  the  indicated  root. 

E.  g.    The  following  surds  are  all  in  simplest  form : 
y'S,     y^a,     ^/(■i^^     VC**  +  ^Y' 
The  following  surds  are  not  in  simplest  form  : 

VI,    ^,    '^^,    Vi8- 

For,  in  ^^\  the  radicand  f  is  not  integral. 

In  V^  the  radicand  8  is  a  power,  8  =  2^,  of  which  the  exponent  3  is  a 
factor  of  the  index  6  of  the  radical.     We  have  v^8  =  ^/^. 

The  radicand  x^  of  '^x*  may  be  expressed  as  a  power,  (x^)^,  of  which  the 
exponent  2  is  a  factor  of  the  index  8  of  the  radical,  and  we  have  Vit®  ^  '\/x^. 

The  radicand  of  the  surd  ^/\'^  contains  as  a  factor  9,  the  square  root  of 
which  can  be  extracted. 

We  have     VTs  =  \/9^  =  3^/21     (See  Chap.  XVIII.  §  20.) 

I.    The  Kadicanrt  an  Integer. 

24.  When  the  radicand  is  an  integer,  the  reduction  of  a  surd  to 
simplest  form  depends  upon  the  following  principles  : 

^57^  =  ^a^/h.     (See  Chap.  XVIII.  §  20.) 

Hence  the  following : 

Separate  the  expression  under  the  radical  sign  into  two  groups  of 
factors^  one  group  containing  all  of  the  factors  the  exponents  of  which 
are  equal  to  or  multiples  of  the  index  of  the  required  root^  and  the 
second  group  containing  all  factors  of  lower  degree. 

Extract  the  indicated  root  of  the  first  group  of  factors  and  multiply 
the  result  by  the  coefficient  of  the  given  surd,  if  there  he  one,  and  write 
the  product  as  the  coefficient  of  the  indicated  root  of  the  remaining 
second  group  of  factors* 

Ex.  1.  Reduce  ^a%H  to  simplest  form. 

We  may  separate  the  radicaud  into  the  two  groups  of  factors,  a%'^  and  he. 
The  first  group  consists  of  all  of  the  powers  of  highest  degree  of  which  the 
exponents  are  exactly  divisible  by  the  index  2  of  the  indicated  root,  and  the 

27 


418  FIRST  COURSE  IN  ALGEBRA 

second  group  contains  all  of  the  remaining  factors  of  degree  lower  than 
the  second. 

Hence,  V^^p^  =  V{.a%*){bc) 

=  v«^v^ 

=  ab^\/l)c. 
In  practice  we  may  omit  writing  the  step  in  the  second  line  above.     It  is 
given  here  simply  to  show  that  the  expression  in  the  third  line  results  from 
the  expression  in  the  first  line  by  applying  the  law 

Ex.  2.   Simplify  5\/TT. 

We  have,  5 ^/li  =  5^4^  =  5 ^^a/^  =  10^3. 

It  is  sometimes  convenient  to  write  the  prime  factors  of  the 
expression  under  the  radical  sign  before  attempting  to  express  it  as  a 
product  of  two  factors  or  groups  of  factors. 

Ex.  3.  ^^^4320  =  4^2^  •  3«  •  5 

=  >^(2«  •  38)(22  •  5) 
=  ^(^^^)\/¥^ 
=  2  •  ^\/W^ 
=  6  >y^20. 

Mental  Exercise  XX.      1 
Reduce  each  of  the  following  surds  to  simplest  form  : 

1.  Vs.  12.   a/45.  23.  a/75.  34.  7a/200. 

2.  Vu,  13.   V^l.  24.  VT25.  35.   V^16. 

3.  a/20.  14.  a/90.  25.  3a/50.  36.  V^24. 

4.  a/28.  15.  a/99.  26.  a/175.  37.  ^32. 

5.  a/40.  16.  2a/27.  27.  4a/250.  38.  6V^40. 

6.  a/52.  17.  3 a/63.  28.  Vl2.  39.  3^48. 

7.  a/60.  18.  a/32.  29.  a/iOS.  40.  5^56. 

8.  3a/24.  19.  a/48.  30.  2a/180.  41.  4^72. 

9.  5A/i4.  20.  a/80.  31.  3a/98.  42.  7^80. 

10.  6a/56.     21.  2a/96.     32.  5a/128.    43.  8^/88. 

11.  a/18.      22.  3a/160.    33.  6a/162.    44.  \/54. 


SURDS 


419 


45.  4V^81.            55.  3V^25(). 

65.  Va^               75.  xVyh\ 

46.  2V^108.         56.  5^600. 

66.  V6^              76.  mVm^n. 

47.  v^32.              57.  9'^128. 

67.  Vc^.               77.  V^V. 

48.  5v^80.            58.  6^/500. 

68.  aVb\            78.  V6y. 

49.  3v'96.           59.  12V700. 

69.  6Vc^             79.  VwV. 

50    \/64.              60.  5V243. 

70.  cVc«.             80.    Va^b^c\ 

51.  a/96.              61.  4V343. 

71.  (?V^^            81.   Va^y^^. 

52.  3V242.          62.  2a/512. 

72.  Vx^f'           82.  Vai°^«c^ 

53.  4a/288.          63.  3V450. 

73.  Vab\            83.  xVx'ijz^ 

54.   lOVlOOO.      64.  3^375. 

74.  VcV.            84.   V4a^6. 

85.  V9cV. 

100.  V(«  +  bye. 

86.  3Vl6A*F. 

101.   A/(iK  -  ?/)  Vw. 

87.  5cVSab\ 

102.  a/(w2  +  nfmn. 

88.  SbVl^c^d*. 

103.   V(b-cybh\ 

89.  6cV24a=^6V. 

104.   V«c''  +  ^f". 

90.  6«A/28a^V. 

105.   Vmif  —  mf, 

91.  3v'8a*6. 

106.   'v/«'"6. 

92.  5'V^40a«^V. 

107.  v^^/n 

93.  4arV27a*6»c2. 

108.  "■  v'c'-. 

94.  2i/V5QxYz\ 

109.  "1i/«"+^. 

95.  3cv^48a^^>V. 

110.  "v^6«-\ 

96.  5  5^81rVz^\ 

111.  a/c^^V. 

97.  2</mg%'k\ 

112.   A/a;2y'*;2;. 

98.  3^v^64rt*^c^ 

113.  ""V«"~'"^',  ?^  >^. 

99.  4gA^243  7?zV/. 

114.   'v/«'-''<^'-'V+«,  r  >  3. 

II.    The  Raclicand  a  Fraction. 

25.  To  reduce  a  surd  to  simplest  form  when  a  fraction  appears 
under  the  radical  sign  : 

Multiply  both  terms  of  the  fractional  radicand  by  the  number  or 
ecppression  of  lowest  degree  that  will  make  the  denominator  a  power 
of  which  the  exponent  is  equal  to  or  a  multiple  of  the  index  of  the 
indicated  root. 


420  FIRST  COURSE  IN  ALGEBRA 

Express  the  transformed  radicand  as  the  product  of  two  groups 
qf  factcyrs^  one  of  which  contains  the  denominator  of  the  tran^ormed 
radicand  and  also  allfactoi's  of  the  numerator  which  have  exponents 
which  are  equal  to  or  a  multiple  of  the  index  of  the  indicated  root ; 
the  second  group  contains  all  factors  of  the  numerator  of  which  the 
exponents  are  not  equal  to  or  multiples  of  the  index  of  the  indicated 
root. 

Extract  the  indicated  root  of  the  first  group  of  factors  and  multiply 
the  result  by  the  coefficient  of  the  given  surdy  if  there  he  one^  and  write 
the  product^  reduced  to  simplest  form^  as  the  coefficient  of  the  indicated 
root  of  the  remaining  second  group  of  factors. 


Ex.  1.   Simplify  |/?. 


"We  may  obtain  a  fraction  the  denominator  of  which  is  the  square  of  a 
rational  number  by  multiplying  the  numerator  and  denominator  of  2/  3  by  3. 

Hence  we  have,  /?  =  /?  =  -^  =  V| . 

Ex.2.   Simplify  i/|^. 
'    8  0 

The  radicand  9a^/Sb  may  be  transformed  into  an  equivalent  fraction  of 
which  the  denominator  is  the  square  of  a  rational  expression  by  multiplying 
both  numerator  and  denominator  by  2  b. 

Hence  we  have,  f  ^  =  f  "l^^  =     ^j^p     =46^2^' 

Exercise  XX.    2 
Reduce  each  of  the  following  surds  to  simplest  form  : 


1.  Vi. 

8.   VtV- 

15.  Wh 

22.  3^. 

2.  VI 

9.  VtV. 

16.  W^j- 

23.  5^S- 

3.  Vh 

10.  30V^. 

17.   VI 

24.  W^. 

4.  bVh 

11.  28  VS- 

18.   Vh 

25.  iVh 

5.  Wh 

12.  Wl 

19.  VI 

26.  tV^tV- 

6.  IIV^. 

13.   Wl 

20.   Vh 

27.   UV^^. 

7.  VI 

u.  Wh 

21.   Vh 

28.  ^Vh 

SURDS  421 


:!:^-  -\/i-  "■%  -v^v 

33.^.  52.^1.         66.  V/^.  80.  y/^^- 

34.   Vf  ,/I  /^■^  .  /T 

o.      /-T  53.  ^>V-.         67.  V--  81.  mV/  —  . 

36.  V^.  54.  V^.         68.  yf.  82.  «6y/^. 

37.  VM'  /-  4  /-^ 

3aiV¥.    '^-bfr  ^^-v^l'     «^-wl;_ 

3«-^i-  56.i^i.'    70.v/l.  84.  V-- 

u.^w.    ^^.41  ^^V|-    «^-M»- 

"•\;?-  58.i;/j.        .2.^71-  86.  Vf- 

43.  V^.  ^'^  ^  '^  c^  6t  V  6 

-^^  -\/i-  "VI-   «^-^:v/f- 

46.4.^A.  ''-V^-  '*•%  ««-V27W^- 

•\/J-       «i-v^i-    ^^V^     ««V8^- 

V^3_63VV"-       77.^-  91.V«3 

92.  ^V/^     94.  (.  +  .)\/|^-   96.  («-^Vf3|- 

93.  («»- V^,-   95.  |i|v/:-^-      97.  («  +  V^V 


47 

48 
49 


422  FIRST  COURSE  IN  ALGEBRA 

98 


99 
00. 
01. 
02. 


.  -i-rV/^-     106.  {/-„'  ^  >'^-      114.  V^^- 

v?->^-  "oVj:.      1.8.  ^{/| 


m-i 
03 


04.^7^-  112-  V^^    .  120.  (/L,  «<8 


05.^1,,.»>2.     1.3.^.  121.  I^i, 


11.  Recliietion  of  a  surd  to  an  equivalent  surd  of  lower 
order,  when  the  index  of  the  root  and  the  exponent  of 
the  radieand  have  a  factor  in  common. 

26.   The  reduction  depends  upon  the  following 

Principle:  TTie  valtie  of  a  surd  remains  unaltered  if  the  index  of 
the  root  and  the  exponent  of  the  radieand  he  both  multiplied  hy  or 
divided  by  the  same  number. 


"^^  =  y/^aP"  =  -C/^.    (See  Chap.  XVII.  §§  19,  21.) 

Ex.  1.    Reduce  ^/s  to  an  equivalent  surd  of  lower  order. 
Expressing  the  radieand  8  as  a  power,  2^,  we  have, 

Ex.  2.         Reduce  'V^25  a%^  to  simplest  form. 


SURDS  423 


Mental  Exercise  XX.   3 
Simplify  each  of  the  following  : 


1.  A^25. 

2.  ^/36. 

14.  ^100  a=^. 

15.  ^/a'b^\ 

27.  W- 

28.  'V^. 

-  {/I- 

3.  n. 

4.  ^9. 

16.   ^a«6«. 

17.  \/xyz^ 

29.  Va5"*r. 

30.  '^x'Y'. 

-  i/J- 

5.  ^27. 

6.  v^l6. 

18.  -v/ieaV. 

19.   v^a«/>V. 

-  yi 

7.   ^a«6». 

20.   ^/xyz\ 

4    / — 

Sn  / — 

8.    A^«^. 

21.   v^wzV. 

^^-  v/f 

38.  Vi-.- 

9.   v'iya^ 

22.  V^. 

»      U 

10.  ^/xY. 

23.  'a//7. 

24.  V^^. 

-  (^1^ 

39.  \/4- 

11.  -^125  a». 

12.  <^a%*c\ 

13.  v'a^/. 

25.  '^?. 

26.  'Vy'. 

-  ^i- 

*o-  V'^^"' 

Addition  and  Subtraction  of  Surds 

27.  Two  surds  are  said  to  be  similar  or  like  if  they  can  be 
expressed  as  rational  multiples  of  the  same  monomial  surd. 

E.  g.  "v/s  and  \/2  are  similar  surds,  for  \/8  may  be  expressed  as  a 
multiple  of  \/2j  as  follows  :   \/s  =  2\/2. 

V  27  and  '\/48  are  similar  surds,  since  each  may  be  expressed  as  a  mul- 
tiple of  \/3,  as  follows : 

Vs?  =  3 a/3,  and  ^48  =  4\/3. 

V  75  and  y^  are  similar  surds,  since  a/75  =  5  \/'S,  and  a/!  =  3  V  3. 

Two  surds  are  said  to  be  dissimilar  or  unlike  if  they  cannot 
be  expressed  as  rational  multiples  of  the  same  monomial  surd. 

E.  g.    y  2  and  Vs  are  dissimilar  surds. 


424  FIRST  COURSE   IN   ALGEBRA 

28.  To  add  or  subtract  like  surdsj  first  reduce  them  to  simplest 
form  and  then  find  the  algebraic  sum  or  difference  of  the  coefficients 
as  a  coefficient  of  the  common  surdfact(yr, 

Ex.  1.   Find  the  sum  of  ^'U.,  -v/3  and  ISyT/S. 
Reducing  the  surds  to  simplest  form,  we  have, 

^12  +  VS  +  18y/i  =  2^3  +  V3  +  6  V3 

=  9^3. 

29.  Unlike  surds  cannot  be  expressed  as  a  single  surd  by  addition 
or  subtraction. 

Ex.  2.    Simplify  2^48  +  5^/20  -  Z^YJz  +  y^- 

2^48  +  5^20  -  ^\\  +  V45  =  8^3  +  10^5  -  V^  +  3  V^ 

=  7V3  +  13V5. 

Exercise  XX.    4 
Simplify  each  of  the  following  : 

1.  V3  +  4V3.  17.  V2  +  V^. 

2.  2\/5  +  3\/5.  18.  a/3  -  Vi- 

3.  4^/13  -  7  Vis.  19.  Vl  -  V6. 

4.  V3  +  Vl2.  20.  Vl  +  a/|. 

5.  V2  +  Vl8.  21.  VI  -  Vf . 

6.  2\/6  4-  \/24.  22.  \/lO  +  VJo  +  V90. 

7.  4V3  +  2\/48.  23.  V2  +  V50  -  ^72. 

8.  V45  -  2  V5.  24.  \/6  —  V24  +  V54. 

9.  2V72  -  5V2.  25.  2^2  +  Vl8  +  V32. 

10.  a/50  +  a/8.  26.  4a/3  +  2A/i2  +  a/75. 

11.  3a/80  -  2 a/20.  27.  2a/Ti  +  5\/44  -  3 a/99. 

12.  5a/72  -  2a/32.  28.  a/|  +  a/3  +  a/^. 

13.  3^16  +  V^54.  29.  A/f  +  A^f  +  ^a/G. 

14.  5^108  -  ^32.  30.  2a/S  +  3a/^. 

15.  5^2  -  aJ^32.  31.  aA/2  +  ^a/2. 

16.  A^2  +  AJ^Slg.  32.  CA/i<^  -  c?v^. 


33.  a\/m  +  Vrn. 

34.  V4^  +  V9^. 

35.  Vl^  —  a/367' 

36.  2 V64^  -  VSui. 

37.  V^  +  Vo^ 

38.  ^/a  +  V^. 

39.  ^'?-c^/~c. 


40.  3a/81  wz  +  2 a/25  w. 

41.  16a/9«  —  9A/l6a. 

42.  A^8^  +  A^. 

43.  aA/rt"  +  a/«"^ 

44.  m'\/m^  +  V?w'. 

45.  a/^  +  a/^. 

46.  a\/Wc  +  a/c. 
47. 
48. 
49. 

50. 


vf 


+  a/2. 


63.  ^a;''j/  + 


vl-v? 


SURDS 


425 


1 


51. 

52. 
53. 
54. 
55. 

56. 

57. 


5V^--cV5. 


Via  +  bf  +  a/^^  +  /^. 

Vx  —  y  —  Vi-^  —  yf- 

a/2  +  A/2a'^  +  a/86^. 
A/a  —  2A/a*  +  VaK 

y/l  +  y/ij  +  y'i:. 
▼  a       ▼  rt^       ▼  a^ 


58.iV^6-v/|4-v/i 


J^  +  J^+J. 

1  1IZ      y  zx      T  ' 


a6 


59. 

60. 
61. 
62. 


64.  2a/^+ 3a/«^  + 4a/^+ 5> 

65.  3\/7  -  a/28  +  a/40  +  a/90. 

66.  2a/6  -  a/54  +  Vu  —  Vu. 

67.  Va+  Vb+  a/o"  +  a/^. 

68.  a/^  —  a/^^  +  a/^  +  a/^- 


aV^b  +  a/4^  +  a/^. 


'b\ 


Reduction  to  Equivalent  Forms 

30.   Any  finite  rational  number  can  be  expressed  in  the  form  of  a 
surd  of  any  desired  order  by  applying  the  law  a  = 


426  FIRST  COURSE   IN  ALGEBRA 

Raise  the  given  number  to  a  power  of  which  the  exponent  is  equal 
to  the  07'der  of  the  required  radical  and^  using  this  result  as  a  radi- 
cand^  express  the  requii-ed  root. 

Ex.  1.   Express  3  as  a  surd  of  the  f(furth  order. 

We  have  3  =  v^3*  =  ^^. 

31.  The  coefficient  of  a  surd  rnay  he  placed  under  the  radical  sign 
and  made  to  appear  as  a  factor  of  the  radicand  by  raising  it  to  a 
power  the  exponent  of  which  is  equal  to  the  index  of  the  indicated 
root. 

This  is  an  application  of  the  principle  a  =  '^a'. 
We  have  ay/h  =  ^a^^Tb  =  V«'*» 

32.  A  surd  is  said  to  be  entire  if  its  coefficient  is  unity. 
E.  g.    V2,  a/^j  V^>  ^"^  entire  surds. 

33.  A  surd  is  said  to  be  mixed  if  its  coefficient  is  a  number 
different  from  unity. 

E.  g.   3'v/5»  o'^J^ft,  (//I  +  w)\An~— "'^j  are  mixed  surds. 
Ex.  2.    Express  the  mixed  surds  2\/3,  3\/2»  and  \^/^  as  entire  surds. 
We  have  2^/Z=  ^/T^   =  y^, 

3V2=  \/9^    =  VI8, 

Mental  Exercise  XX.    5 
Reduce  to  the  forms  of  surds  of  the  orders  indicated  : 

1.  2,  2nd  order.  5.  6,  3rd  order.  9.  ^,  5th  order. 

2.  3,  4th  order.  6.  2,  5th  order.  10.  a,  6th  order. 

3.  h,  4th  order.  7.  12,  3rd  order.  11.  ->  5th  order. 

4.  I,  3rd  order.  8.  4,  4th  order.  12.  a"",  nth  order. 
Transform  each  of  the  following  mixed  surds  into  an  entire  surd  : 

13.  2V3.  15.  4V2.  17.  S\^.  19.  4v^3. 

14.  3V5.  16.  2'v/2.  18.  5\/2.  20.  6^^. 


21.  iV2. 

34.  «V^. 

22.  iVe. 

35.  bVc. 

23.  iVS. 

36.  cv^^. 

24.  iVl- 

37.  m'^n. 

25.  3V|. 

38.  aVS. 

26.  5VS. 

39.  2^v'y. 

27.  iVV5. 

40.  ahyj~c. 

28.  Wl 

41.  ar^Vi^. 

29.  iVi 

42.  JcV^^. 

30.  3V|. 

43.  VS 

31.  Wl 

^v^. 

32.  i^i. 

33.  fv^f. 

44.  y.. 

SURDS  427 

45.  iv/^.  51.  ^V^. 


y 

n 


46.  Vi-  ''■  W 


54.  a^/h. 
48.  iyi.  55.  /^v^c. 

49 


56.  c\/c. 

57.  ^a/^. 

58.  x'y^x. 

59.  icv"^. 
^V  3              60.  a^'^/a. 


2V  a 


«■  Iv/  ■ 


Change  of  Order. 

34.  Surds  of  different  orders  can  be  transformed  into  equivalent 
surds  of  the  same  order  by  applying  the  principle  ^/a''  =  Vo^"* 
(See  §  26). 

Using  radical  symbols,  we  may  proceed  as  follows  : 
As  a  common  index  fm^  all  of  the  transformed  surds^  write  the 
lowest  common  multiple  of  all  of  the  indices  of  the  given  indicated 
roots.  Then  raise  each  radicand  to  a  power  the  exponent  of  which 
is  equal  to  the  number  by  which  the  root  index  must  be  multiplied 
to  produce  the  lowest  common  multiple  of  the  indices. 

Ex.  1.  Transform  -y^S,  ^^2  and  ^y/S  into  equivalent  surds  of  the  same 
order. 

The  lowest  common  multiple  of  the  indices  2,  3,  and  6  is  6. 

Hence,  ^3  =  v^3^  =  ^^27, 

^2  =  ^2"2  =  ^\ 

35.  When  transforming  surds  of  different  orders  into  equivalent 
surds  of  the  same  order  it  is  often  convenient  to  use  the  notation 
of  fractional  exponents,  and  to  proceed  as  follows  : 


428  FIRST  COURSE  IN  ALGEBRA 

Express  each  of  the  indicated  roots  by  using  the  notation  of  frac- 
tional exponents,  BedvLce  the  fractional  expmients  thus  obtained  to 
equivalent  fractional  exponents  having  a  lowest  common  denomina- 
tor. Express  the  results  thus  obtained  in  the  radical  notation^  observ- 
ing that  the  numerator  of  the  fractional  exponent  denotes  the  power 
to  which  the  radicand  is  to  be  raised  and  that  the  dmwminator  is  the 
index  of  the  required  root, 

Ex.  2.   Which  is  the  greater,  y/l  or  ^1  ? 
We  have  ^5  =  5^  =  52^  =  ^5*  =  ^625 ; 

also,  ^  =  7^  =  7^=  ^7^=  ^343. 

Since  625  >  343,  it  follows  that  ^625  >  ^343. 
Hence,  ^b  >  ^. 

Exercise  XX.    6 
Express  as  equivalent  surds  of  the  same  order  : 

1.  a/2  and  \^5.  7.  v^8  and  ^l.  13.  Va  and  ^\ 

2.  V3  and  v^.  8.  ^2  and  </l.  14.  -v^^^and  ^/b\ 

3.  V5  and  V^,  9.  V3  and  -^15.  15.  Vx  and  v^^. 
*4.  V6  and  \^U.  10.  V^  and  '^80.  16.  Vy  and  "-C^y. 

5.  V7  and  \/^.        11.  Vs  and  -^75.        17.  ^/i^and  a/^^ 

6.  '^2andA!^3.  12.  V2  and  ^^12.        18.   'fwand"^?^. 

Which  is  the  greater, 

19.  3\/5or4\/3?  25.  v^S  or  v^?  31.  a/6  or  v"^? 

20.  4a/6  or  5a/3  ?  26.  a/5  or  v^20?  32.  2  or  a^7  ? 

21.  6A/2or2A/61  27.   A/6or2A^2l  33.  3  or  a/30  ? 

22.  5 a/7  or  8 a/3?  28.  \/l  or  a^50?  34.  4  or  2a^3  ? 

23.  A^or2v^?  29.  V^  or  2v^  ?  35.  5  or  3a^7  ? 

24.  3v^  or  2'^T0 1  30.   a/3  or  a^ ?  36.  3  or  2'V^IO  % 
37.  Va  or  a^^  for  a  >  1  ?       38.   a/^  or  v^«,  for  b  >  I. 

Multiplication  of  Surds 
36.   The  product  of  two  monomial  surds  of  equal  orders  may  be 
found  by  applying  the  principle  ^s/ay^b  =  Vab.     (See  §  24.) 


SUKDS  429 

Xf  the  given  surds  are  of  different  orders^  they  must  first  he  trans- 
formed into  equivalent  surds  of  the  same  order. 

Multiply  the  coefficients  together  for  a  new  coefficient^  and  the 
expressions  under  the  radical  signs  for  a  new  7'adicandj  and  reduce 
the  result  to  simplest  form. 

Ex.  1.     5/v/6  X  3v^  =  5  •  Sy'e^  =  15 y^  =  SOy^S. 

Ex.  2.     5  V^  X  v^2  =  ^5»  X  '^¥^  =  '^58  x  2^  =  ^500. 

It  is  often  convenient  to  obtain  the  prime  factoi-s  of  the  radicands  before 
multiplying. 


Ex.  3.     V35  X  V^l  =  V^  •  7  x  \/V^  •  7  =  ^^1^  •  5  •  13  =  7V65. 

Exercise  XX.     7 
Write  each  of  the  following  products  in  simplest  form  : 

1.  a/8  X  \/5.  20.  v^«6V  X  ^/¥ed. 

2.  ^2  X  V6.  21.   ^xyh^w^  X  v^?A«^. 

3.  V5a  X  VlOa.  22.  Vi  X  Vi- 

4.  VS^  X  Vl5.  23.  Vt  X  Vf . 

5.  Ve  ^  X  a/12^c.  24.  V?^  X  Vv^. 

6.  \/\^  X  ^3^.  25.  4\/V  X  V¥. 

7.  v^4^  X  v^8  ar/. 

8.  '^S^  X  '^16^.  26.  y/^  X  y/^;. 

9.  5a/15  X  V^- 

10.  2a/14c  X  3VT^. 

11.  6Vl2  X  4a/8^. 

12.  SVTS  X  9\/2(). 

13.  ^45  X  ^18.  28.  i/^  X  \/- 

14.  v^X  A/6aa;.  ^  ^       ^^ 

15.  ^^14  X  ^^21. 


27 


Y!x# 


16.    ^ 


17.   v^tcy^^;  X  ^Jxy^zK 


18.  ^/'-la^hc'^y.  \/ia^b^c. 

19.  -v/oP?  X  <^Mi. 


30 


430  FIRST  COURSE  IN  ALGEBRA 

31.  i/I  X  v/^.  ^^-  '^'^^  '^^'^^' 

'  y  be      ^  ab  .Q7    ^aA**-!  V  a/^ 


32 


v/^x{/Z. 


37.  wv'w^'  X  -</;?. 

38.  v^  X  ^V^SO. 
^«                             39.  V^  X  V55. 

33.  v^a  X  v^a^.  40.  Vf  X  ^J. 

34.  V^^^  X  ^^lr^\  41.  4  a!^  X  6^. 

35.  Vd^  X  \^.  42.  3^^  X  9</9^. 

Multiplication  of  Polynomials  Involving  Surds 

37.  The  product  of  two  polynomials  the  terms  of  which  contain 
surds  may  be  obtained  as  follows : 

Multiply  each  term  of  the  multiplicand  by  each  term   of  the 
multiplier. 

Ex.  1.  (V^-5/y/3  +  'v/6-7)x3V6=3\/l2-15'v/T8+3\/36-21V'6 

=  6y^3  -  45^2  +  18  -  2iy'6 
=  18  -  45v^  +  6>v/3  -  21/v/6. 

Ex.  2.    Multiply  by/b  +    2^/^  by  4^/5  -  3y^. 

5v^+    2^2 
4<v/5-    3v^ 

100+    8/v/^ 

-15^10-12 

100-    7^10-12  =  88-7/^. 

38.  Two  binomial  quadratic  surds  which  differ  only  in  the  sign 
of  one  of  the  surd  terms  are  called  conjugate  surds. 

E.  g.  'y/ft  +  V^ft      aud     ^/a  —  ^h ;    5  +  4-^/3      and      5  —  4y'3  ; 

-v/6  +  V^      and  -  ^6  +  ^1. 

39.  T^  product  of  two  conjugate  surds  is  a  rational  number  or 


Representing  two  conjugate  surds  hj  ^/x  +  y^    and    ^/x  —  y\/y,   we 
have,    (yx  +  ^/y){^/x  -  ^y)  =  (\^x)^  -  {^/yY  =  x-y. 

Ex.  3.         (2^7  +  ^/U){2^1  -  yU)  =  28  -  11  =  17. 


SURDS  431 

Exercise  XX.     8 
Simplify  each  of  the  following  expressions  : 

[Vn  -  Vl3  +  V7)a/3.  9.   (3a/2  -  2\/3  +  ^12)^6. 

y2  +  Vl  —  V3)V5.  10.   (V5  -  V2I  +  4^/27)2^3. 

;a/7  -  A/Tl  4-  Vl3)V6.        11.   (2a/5-4\/10-V30)3V5. 
;Y^  -  VlO  -  VT5)a/7.       12.  (V«  +  Vb+  Vc)Vabc., 

(\/2  —  V5  4-  Vio) a/To.     13.  ( V^  +  a/^  +  Vzx)Vx^- 

;\/2  +  Vt  +  2a/14)aA4.      14.   (a^  +  \^  -  '^10)^. 
;a/3  -  a/5  +  a/10)a/15.        15.   (^9  -  2</s  -  ^18)'V^3. 
yiO  +  2a/3  -  a/5)a/5.        16.   (^4  +  ^9  +  'V^SG) V^G. 
y  100  -  'v/25  +  V^4)^10. 

[^;^c  -  ^/Wd  +  ^/'(?d)^/abdi. 

y^  +  A^^  +  a/0\^«^- 

;V3  +  4)  (a/2  +  3). 

;2  4-  V3)(3  -  a/2). 

[a/6_-  5)(V_3  +  5). 

^2a/7  +  7a/2)(3'v/7  +  8a/2). 

;10a/G  —  6a/T0)(5a/3  +  3a/5). 

^2  +  '^3)(^4  +  A^5). 

:i  +  a/2  -  a/3)(a/2  -  a/6). 

V3  4-  a/5  X  V^s  -  Vs. 

V^6  4-  3a/3  X  V^6  -  3a/3. 

a/2  —  a/3  4-  a/5)(a/2  4-  a/3  -  a/5). 
a4  -  a/|  4-  A/i)(A/6  4-  a/7  -  a/8). 

W\  -  A/f  -  V^)(a/§  -  a/^  -  a/?. 

a/2  -  a/3)(a/3  -  a/5)(a/5  -  \/l). 

'a\/h  —  ^sfaJ)  +  h^/a){Va  —  ^b). 

^ah  +  \/bc  +  ^fm)(y'\/a  +  a/^  4-  a/c)- 

A/a  4-  a/^  +  a/c)(a/«  4-  a/^  —  a/^)(a/^  —  a/^  4-  a/c) 

(a/^  —  a/^  —  a/c). 


1 


432  FIRST  COURSE  IN  ALGEBRA 

Involution  of  Monomial  Surds 

40.  From  the  principle  {^/af  =  'C/a",  it  follows  that : 

An  entire  monomial  surd  may  be  raised  to  a  power  by  raising  the 
radicand  to  the  indicated  power. 

Ex.  1.       {^ly  =  >v^  =  b^Z. 

Ex.  2.    (2 'v/eTt)'  =  2V6*a*  =  48 a  ^/Wa. 

Ex.  3.    (^5^)2  =  (5 a)^  =  (5 a)^  =  ^u 

Mental  Exercise  XX.    9 
Reduce  the  following  indicated  powers  to  simplest  form  : 

*  1.  (\^2y.  13.  (V7t)\  25.  (c'^Viy. 

2.  (\^5y.  14.  (Vf^y.  26.  (c\^y. 

3.  (\^sy.  15.  (V^y.  27.  (d^ny. 

4.  (V2)«.  16.  (V?)l  28.  (a\^cy. 

5.  (V5)*.  17.  (\/c)^  29.  (b\/^'y.      •. 

6.  (>^8)l  18.  (v^)*.  30.  (x'\^y. 

7.  (2V3)«.  19.  (\^hy.  31.  (2a;Vi)'. 

8.  (3^/5)*.    .  20.  (v^=^)'-  32.  (^y\^y. 

9.  (4^4)^.  21.  {\/^^f.  33.  (Sa^^i^WO^. 

10.  (2^2)^  22.  (-  ^/J^y.  34.   ('v/«)^   7^  >  2. 

11.  (-2V7)'.  23.  (aVa)'.  35.  ('^/^)^   7^  >  3. 

12.  (V^^^)*.  24.  (-  b^/yy.  36.  (a;A/y)^    w  >  4. 

Division  of  Surds 

41.  The  quotient  obtained  by  dividing  one  entire  monomial  surd 
by  another  may  be  found  by  applying  the  principle 


The  process  for  finding  the  quotient  of  one  mixed  surd  divided  by 
another  may  be  made  to  depend  upon  the  principle  above. 

Since  two   surds   of  different  orders  can  be   transformed  into 


SURDS  433 

equivalent  surds  of  the  same  order,  it  follows  that  it  is  necessary  to 
state  the  process  only  for  mixed  surds  of  the  same  order. 

The  index  of  the  indicated  root  of  the  radicand  of  the  quotient  ob- 
tained by  dividing  one  monomial  surd  by  another  of  the  same  m'der  is 
equal  to  the  common  index  of  the  indicated  roots  of  the  radicands  of 
the  dividend  and  divisor. 

For  the  coefficient  of  the  radical  part  of  the  quotient  divide  the 
coefficient  of  the  dividend  by  the  coefficient  of  the  divisor,  and  for  the 
radicand  of  the  quotient  divide  tJie  radicand  of  the  dividend  by 
the  radicand  of  tlie  divisor. 

The  result  should  be  reduced  to  simplest  form, 

Ex.  1.    Sz/Il  -r  4V^  =  (8  ^  A)^f\A^  =  2^7. 
Ex.  2.   Divide  21^2  by  7^. 

We  have,  "^  =  -^M=.  =  3  JZ?!  .  3;/III^ 
7^Q       V^(2  •  3)2        V  22  .32        V  32  .  34 


=  a7162. 


Exercise  XX.     10 

Simplify  each  of  the  following  quotients  : 

1.  V6  -T-  a/2.  15.  ^^27  -^  4^3.  29.  5^2  -^  \/lO. 

2.  VlO  -^  V5.  16.   ^50  -r-  ^2.  30.  3V5  H-  Vl5. 

3.  a/14  ^  a/2.  17.   '^in  -^  ^6.  31.  6a/7 -^  ^42- 

4.  V'2T  -7-  \/3.  18.   \/7  -7-  a/2.  32.  10a/3  -t-  VQ- 

5.  a/15  H-  V5.  19..  a/3  -^  Vs.  33.  14a/2  -^  VH. 

6.  3V22  ~  a/Ti.  20.   VTO  -^  V3.  34.  a/7  -f-  3a/21. 

7.  5a/26  ^  a/T3.  21.   a/13  -t-  a/7.  35.  Vn  -i-  2a/22. 

8.  8a/30  H-  2\/2.  22.  a/10  ~t-  a/6.  36.  V3  -^  4a/15. 

9.  a/39  -f-  2a/13.  23.   VTs  -^  a/To.  37.  a/2  -i-  5a/6. 

10.  a/sT  -7-  3a/17.  24.  A/2I  ^  a/14.  38.  A^a^  -^  a/^- 

11.  Vi  -r-  V2.  25.   A^2  -  Vi.  39.  V6^-  Vc._ 

12.  V9  -T-  Vs.  26.   VlO  -^  VT2.  40.  ^Mc  -f-  Va^'. 
1.3.  VT2  -^  Vi.  27.  Vs  -r-  V2.  41.  V^  -^  ^v^. 
14.  5  V18  -T-  Va  28.  V5  -^  Vi.  42.  Va^  -^  Va. 


434  FIRST  COURSE   IN  ALGEBRA 

43.  ^^  -^  ^.  48.  ^  ^  y/h\  53.  ^/xyz  -f-  xy^fz, 

44.  ^/W?^^/Wc.  49.  v^^v^.  54.  ah^Tc  ^  a^/Vc. 

45.  Va  -T-  V^»  50.  av^  -j-  Va.  55.  Vicj^-s  -r-  "s/yzw. 

46.  Vc^Vx.  51.  6a/^  -i-  \/6.  56.  ^i^  -r-  V^^^. 

47.  Vrf  -^  a/^.  52.  av^  -r-  6\/a.  57.  V^  "^  V^. 

58.  V2c  -^  V3^.  62.  aJ/^  -^  v^a,   ^^  >  2. 

59.  V7« -H  A/l4y.  63.  v'^*^  -h  v^^Za^,    w  >  1. 

60.  ^lOab  -r-  Vi56c.  64.  (^30  +  a/42)  H-  a/6. 

61.  6A/3«y -T-  3A/6az.  65.  (12 a/35  —  a/45)  -f-  3a/5. 

66.  (8v/5i  -  5^^)  -T-  (  -  2^3). 

67.  (a/15  -  \/6  +  a/2  -  a/3)  -t-  a/3. 

68.  (a/2  +  a/8  —  a/21)  -7-  a/2. 

69.  (8a/7  -  6\/5  +  4a/3)  -i-  4a/2. 

70.  (10a/15  -  5V3  +  3a/5)  -^  30a/15. 

71.  (2A/^-8v^-4)-^^. 

Rationalization 

42.  To  rationalize  a  surd  expression  is  to  free  it  from  indicated 
roots. 

If  the  product  of  two  irrational  factors  is  a  rational  number,  either 
factor  is  called  the  ratioualiziug  factor  of  the  other. 

E.  g.  The  rationalizing  factor  of  y^  is  y  9,  since  /y^3  x  \/9  =  y^27  =  3, 
which  is  a  rational  number. 

43.  From  the  identity  V^  X  'C/a^  =  -^a*"  =  a,  it  may  be  seen 
that  when  p  is  less  than  r  the  rationalizing  factor  of  a  simple  entire 
monomial  surd  represented  by  "^a^  is  A/a'""^ 

Q 

Ex.1.   Rationalize  the  denominator  of — — . 

a/2 

By  multiplying  both  terms  of  the  fraction  3/^/2  hj  /y/2  the  value  of 
the  fraction  remains  unaltered  and  the  denominator  is  made  rational. 

_.,    ,  3        3^2        3^2 

We  have  -— r  =      .J    ,_  =  —~  • 

>v/2      a/2a/2         ^ 


Hence  ^3/3—     Ao.VT  "  "2"  —  ^V^^^* 


SURDS  435 

Ex.  2.   Rationalize  the  denominator  of  — ^~z  • 

^2 

The  rationalizing  factor  of  a/^  is  y^  =  y^. 

^2~     ->^2^4 

44.  From  the  identity  (V«  +  \^~l>){^/a  —  VT>)  =  a  —  b,  it  ap- 
pears that  a  binomial  quadratic  surd  may  be  rationalized  by  multi- 
plying by  the  conjugate  sm^d. 

Ex.  3.  Reduce  (3  +  V^)/(^  ~"  V^J)  to  an  equivalent  fraction  whose 
denominator  is  rational. 

The  rationalizing  factor  of  the  denominator  3  —  /y/5  is  the  conjugate 
surd  3  +  /y/5. 

Hence, 

3  +  V5  _  (3  +  a/5)(3  +  V5)  _  9  +  6^5  +  5  _  14  +  6\/5  _  7  +  3/v/5^ 
3- V5~(3- V5)(3+ V^)~         9-5         "  4  ~         2        * 

Ex.  4.    Divide  y^  +  V^  +  \/6  1^7  V^  +  1. 

Expressing  the  quotient  as  a  fraction,  and  rationalizing  the  divisor,  we 
have, 

V2+V3+V6_ ( V2+v/3+/v/6)(V'3-l)  _ 3+2/^-^^/3 _ 3        r-_\     r- 
VS+I        ~  (V3+l)(\/3-l)        ~  2  "2^2'^* 

45.  From  the  identity, 

(  Va  +  \1  +  Vc)(  Va  —  V6  +  4/c)(  Va  +  v^5  —  Vc)(  i^  —  V6  —  Vc)  =  a*  +6"  \-c'^—2ah—2ac—2he, 

it  appears  that  the  rationalizing  factor  of  any  one  of  the  factors  of 
the  first  member  is  the  product  of  the  remaining  three  factors. 

E.  g.  The  trinomial  surd  (V^  —  \/3  +  ^^5)  may  be  made  rational  by 
multiplying  by  the  product 

(V^+  /y/S  +  //5)(/s/2  +  a/3  -  \/5)(V2  -  a/3  -  V^)- 
When  rationalizing  a  trinomial  surd  it  will  often  be  found  con- 
venient to  group  the  terms  of  the  trinomial  and  regard  the  expres- 
sion as  a  binomial,  of  which  one  of  the  terms  is  a  sum  or  a  difference. 
After  multiplying  by  the  conjugate  binomial  factor,  the  terms  of  the 
resulting  expression  may  be  combined  and  the  process  of  rational- 
ization may  be  again  applied. 


436  FIRST  COURSE  IN  ALGEBRA 

To  rationalize  either  the  denominator  or  the  numerator  of  a  frac- 
tion, multiply  both  numerator  and  d^rwminator  by  the  rationalizing 
factor  of  the  term  to  be  rationalized. 

Ex.  5.    Rationalize  the  denominator  of  '\/3/('\/l0  —  a/Q  +  ^/2i). 
We  have, 

V3 V3(VTo- ye-  yii) 

Vio  -  Ve  +  \/3  ~  (\/io  -  V6  +  V^XVi^  -  V6  -  Va) 

_  ^^30  -  3  ^2  -^  3 
~       13-4  V15 

_  (y^30  -  3V2  -  3)(13  -f  4^15) 
~       (13  -  4Vl5)(13  +  4V15) 
_  >y/3o  -  i2yT5  +  2\^/^  -  39 
-71 

=  f!  -  ?^\/2  +  ^f  Vl^  -  ^r  V30- 
46.   The  object  of  rationalizing  the  denominator  of  a  given  frac- 
tion is  to  avoid  the  use  of  a  divisor  consisting  of  a  non-terminating 
decimal. 

E.  g.   To  find  the  value  of  -^ ,  correct  to  four  places  of  decimals,  if  the 
\/3 
denominator  is  not  rationalized,  it  is  necessary  to  divide  1.41421  +  by 
1.73205 -f,  as  follows: 

V2_  1.41421 +_     V 
:^- 1.73205 +--^1^'  +  - 

While,  by  rationalizing  the  denominator  of  the  fraction,  we  may  obtain 
the  required  value  by  dividing  2.4492  -\-  by  3,  as  follows : 

^3        3  3 

Exercise  XX.   11 
Rationalize  the  denominators  of  each  of  the  following  : 

1  J-.        3  ii-         5  -^-         7  i- 

V6  VT  V^  V^5 

2.  — p-  4.  -7^-  6.  — =•  o.  -37=- 

V5  Vn  a/10  V4 


SURDS  437 

9.-4.  13.-1^.            17.  -A..  21.^. 

10.  -^'  14.    -^.           18.   -^.  22.  ^. 
A/2I                     3V1I                    3'V^8  V12 

11       12  _         7                  .^8  ^..    2a/3 

11.  -^=:^*  15.    — r  19.  ,, ♦  23.     —' 

V28                      5a/14                   9V64  3\/2 

12. -IL.  i6.4-_.      20.^.  24.4.. 

^27                      2V9                       V2     ■  2V6 

25.  ^-^.                   31.  -^.                37.  ^+^. 

V  20                            Vx-y  a/5  -  a/3 


13. 

17. 

2a/5 

14. 

3a/11 

15. 

16. 

'•            20. 
2^9 

a/^ 
31.  ^^^. 

Vx  —  y 

32.     /       . 

A/a;+  2 

33.  /    . 

34.  ''^  +  ^. 

V-2 

35.     /      • 

36.          "      _. 

26.  -^.  32.  -^.  88.  ^^-^. 
a/3+1                           A/a;  +  2  V3  +  a/2 

27.  -^=1—  33.  _^.  39.  4±V^. 
Vll  -  2                         a/»  +  2  a/7  +  V2 

28.  -i^.  34.  ^-4±^.     .  40.  ^. 
■\/3  +  2                              V2  i  +  V2 

29.  iH^.  35.  -^-         41.  ^:$. 

V  5  +  4  yb  —  c       .  a  +  yx 

30.'^.  36.      ^'^        .  42.  ^-f. 

V  3  —  1  a/^  —  A^c  ^x  +  a/^ 

^3    V5+  a/6  ^^    ^vV±rV^. 

a/7  +  a/S  ^a/^J  —  aJA^J/ 

^^    2a/3  -  3a/2  ..     Vo^^  +  5  +  3 

44.  — — -'  48.      , 

3a/6-2a/2  a/«'  +  5  —  3 

45.  2^-^A  49.  ^^ 


9a/8  -  7a/6  a  +  A/a'  -  ^ 

.-    aA/^  +  6v^  _  6 

46.  ;= —'  50. 


a/6  +    a/^  A/a3+  6  —  A^a;  —  6 


438  FIRST  COURSE  IN   ALGEBRA 

51.  V^+v^.  5g_ 


2a/« 

-3- Va 

J  +  6  4-  3  V«  -  b 

3Va 

J  +  6  -  2  Va  -  6 
+  3  +  V<«  —  ii 

+  3- 
+  1  + 

■  V^*— 3 
y/x  4-  2 

V^ 

-1- 

—  1  - 

.  Va?— 2 

Va^ 

-Vic^+1 

52.  -v^^^.  ..v^^^^^  ^^^ 
3Va  +  6  —  2^/a  —  b 

Va  +  3  +  v;r^3      ■ 

53.  — — ■  58. 
V  a  +  3  —  yo-—  3 

54.  ^f^^V^,  59^ 


55.    "^^  ^ ,     ^     '  60. 


1  +  V2  +  V3 
V5- V7 

2  +  V5  +  \/6* 
A/T(")+y2-  a/5 
VIO  —  V2  +  V5 
a/3+  a/5+  a/2 

2a/15  +  6 
ic  +  a  +  A/aJ^  —  «^ 


A/ar*  —  1  +  A/ar^  +1  a;  +  «  -  a/^'-^  -  «' 

47.*  By  means  of  the  identities  in  Chapter  VIII.  §  59  a  ration- 
alizing factor  can  be  found  for  any  binomial  surd  '^x  ±  'Vy. 

(i.)    Vx  —  ^y  can  be  rationalized. 

By  letting  ^x  =  a,  and  '^y  =  b,  and  representing  the  lowest 
common  multiple  ofp  and  q  by  ;*,  it  may  be  seen  that  a"  and  6" 
are  both  rational. 

For  all  values  of  n  we  have, 

(a  —  6)(a"-^  +  a"-2^  + +  a^""*  +  ^""^  =  a"  -  //. 

Hence,  the  rationalizing  factor  of  "^x  —  Vy  may  be  written  by 
referring  to  the  polynomial  factor  iabove. 

Ex.  1.  Find  the  rationalizing  factor  of  a/5  —  a/^- 

The  lowest  common  multiple  of  the  exponents  3  and  2  is  6.  Accord- 
ingly, it  is  necessary  to  multiply  the  binomial  by  such  a  factor  as  will  raise 
both  terms  to  the  sixth  power. 

The  polynomial  rationalizing  factor  may  be  found  as  follows  : 

Eepresenting  /y^  =  5^  by  a  and  \/2  =  2*  by  b,  it  follows  that  the 
binomial  yE  —  a/2  may  be  represented  by  the  binomial  a  —  b. 

If  a  —  6  be  multiplied  by  a^  +  «*6  +  a%^  +  a%^  +  ab^  +  b^  the  product . 
will  be  a®  —  6*. 

Hence  the  rationalizing  factor  of  the  given  binomial  may  be  constructed 
by  substituting  53  for  a  and  2 2  for  6  in  the  polynomial 
a5  _|.  a*jj  ^  a%^  +  a^b^  +  ab*  +  ¥. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


SURDS  439 

Hence  the  rationalizing  factor  of  yS  —  y^  is 

5!  -|_  5I22  +  5^22  +  532t  4_  5^2^  +  2^ 
This  factor  reduces  to  b^Io  +  5^5/y/2  +  10  +  2^)^25 y^  +  -i^  +  4^^. 

(ii.)    U  n  be  even,  'v^  +  y^y  can  be  rationalized,  since 
(a  +  ^)(a""'  -  a""'^  + +  «6"-^  -  ^"-^)  =  a'^  -  ^". 

(iii.)    If  n  be  odd,  '{^ic  +  \^i/  can  be  rationalized,  since 
(a  +  6)(a"-i  -  a"-'6  + -  a^/*^-^  +  b"-^)  =  a"  +  ^>\ 

Exercise  XX.     12 

(This  exercise  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Find  the  rationalizing  factor  of  each  of  the  following  binomial 
surds  : 

1.  1  +  ^2.  6.  VS  +  V^. 

2.  2  4-  V^S.  7.  ^7  -  VTO. 

3.  5  -  V^I  8.  V5  -  \/Q. 

4.  '^5  +  1.  9.  '^G+  v^9, 

5.  VS-\-  ^2.  10.  ^^2- aJ/3. 

Factors  Involving  Surds. 

48.  Extending  the  idea  of  "factor"  to  include  expressions  in 
which  surd  numbers  appear  among  the  coefficients,  we  may,  by 
applying  the  principles  of  Chapter  XII.,  transform  certain  expres- 
sions so  that  they  shall  appear  as  products  of  factors  involving 
surds. 

We  will  now  consider  the  problem  of  factoring  the  general  ex- 
pression of  the  second  degree  containing  one  unknown,  x  : 

ax^  -{-  bx  +  Cj  a  ^  0. 
We  may  write,   ax^  +  bx  +  c^     ax^  -\ i — - 


=  a\  x^  +  -  X  -{-  - 


In  the  trinomial  square  a^  ±  2ab  +  V^  the  third  term  ^^  is  the 
square  of  the  quotient  obtained  by  dividing  the  "  middle  term  "  oir 


440  FIRST  COURSE  IN  ALGEBRA 


"finder  term ''  ±  2ah  by  twice  the  square  root  of  the  first  term  a^. 
(See  Chapter  XII.  §  21.) 

The  process  of  obtaining  a  trinomial  square,  a^  ±  2  aft  +  ft^  by 
adding  the  term  ft'*  to  a  binomial  such  as  a^  ±  2  aft,  is  called  com- 
pleting the  square  with  reference  to  d^  ±  2  ah.  (See  also 
Chapter  XXII.  §§  18-20.) 

We  may  complete  the  square  with  reference  to  the  binomial  x^  + 

- X  which  appears  in  the  expression  a\  x^  -\-  -x-\-  -     as  follows  : 

Dividing  the  "  finder  term  "  -  a;  by  twice  the  square  root  of  the 

first  term  t^  we  obtain  -—  which  is  the  term  whose  square  must  be 
2a  ^ 

added  to  complete  the  square  with  reference  to  a^  +  -  a% 

Hence, 

-[(-A)'-(v/51)'] 
-L(-rJ'-(v/^^yj 


(      ,     h    ^   VW--Iac\f     .     ft       A/ft'-4ac\ 
V         2a  2a        J\        2a  2a        J 

Ex.  1.    Factor  x^  -  10.  Check.     Let  a:  =  1. 

x2  -  10  =  a:2  _  (ylo)2  _  9  =  -  9. 

(See  Chapter  XII.  §  22.)        =  [a:  +  y'lO]  [x  -  y^]. 

Ex.2.    Factoric2  + 6a;  +  4.  Check.     Let  x  =  1. 

11  =  11. 
Using  6  ic  as  a  "  finder  term,"  we  may  complete  the  square  with  reference 
to  a;'^  +  6  a:  as  follows : 

We  have  —  =  3. 

2a; 

Hence,  a:2  +  6x  +  4  =  a;2  +  6a;  + 32-9  +  4 


SURDS  441 

(See  Chap.  XII.  §  18.)  =  (x  +  3)2  -  (^by 

(See  Chap.  XII.  §  22.)  =  [x  +  3  +  ^5]  [x  +  3  ~  y^]. 

Ex.  3.    Factor  3  a;^  +  8  a:  —  5.  Check.     Let  a:  =  1. 

The  following  luethod  may  be  employed  :  6  =  0. 

3a;2  +  8a;-5  =  ^[9x2  +  24a:-  15]. 

The  square  may  be  completed  with  reference  to  9z^  +  24cX  by  using  24  x 
as  a  "  finder  term,"  as  follows : 

We  have,  2S  =  ^• 

Hence,  3a;2  +  8 a;  -  5  =  i [9 a:2  +  24a;  +  4^  -  16  -  15] 

=  H(3^  +  4)2-(y31)2] 
=  |[3x  +  4+  V31][3a:  +  4- V^]- 
Observe  that  in  each  of  the  examples  above  the  factors  obtained 

are  of  the  Jirst  degree  with  reference  to  the  letters  appearing  in  them. 

Exercise  XX.     13 
Obtain  factors  containing  surds  for  each  of  the  following  : 

13.  x^-  Ux-  18. 

14.  9c?^-  12^+  1. 

15.  a^H-  16a+  19. 

16.  b'-l^b-^  57. 

17.  c'-  10  c-  100. 

18.  x'-x-l. 

19.  x^  —  x-\-l. 

20.  x^-'dx-b. 

5        75 

3        72 

Evolution  of  Surds 
49.   A  root  of  a  monomial  surd  may  be  found  by  applying  the 

principles  \  rZ^  "  nr^     (See  Chap.  XVIII.  §  21.) 

(  \Va=  Vva. 

Ex.  1.  \l'^  =  ^1. 

Ex.  2.  V'^SoS  =  Y^^8a8  =  ^2^. 


1. 

a^^-3. 

2. 

x'-^. 

3. 

x"-^. 

4. 

ar^-50. 

5. 

92^2-13. 

6. 

16;2^-5. 

7. 

^"  +  82;+  1. 

8. 

m^  +  6yw  +  2. 

9. 

a^+  12a -3. 

10. 

^>2-106  +  20. 

11. 

r'^  +  2r-  1. 

12. 

m^—  lOm-  17. 

442                        FIRST  COURSE  IN  ALGEBRA 

Exercise  XX.     14 
Simplify  each  of  the  following  : 

1.  V^.  9.  V^8\/8^. 

3.  v^-c^.  11.  Vy^. 

4.  V^VG^.  12.  ^2V% 

5.  n/^^.  13.  21/2V2V2. 

^-  ^^  14.  |y^?P. 

^-  /^^^-  15.  3V^.3?I. 


8.  Va^V^.  16^  >/2W2  2/V2;3«. 

Properties  of  Quadratic  Surds 

50.*  In  the  statements  and  proofs  of  the  following  principles,  the 
radicands  are  restricted  to  positive  commensurable  values. 

(i.)  T/ie  product  or  the  quotient  of  two  similar  quadratic  surds 
is  rational. 

For  if  a,  6,  and  c  be  rational  numbers, 

aVc  X  b\/c  =  ab\/^  =  abc. 

<l\/C  fl 

h^/c  ~~  h 

(ii.)  The  product  or  the  quotient  of  two  dissimilar  qvxidratic  surds 
is  a  quadratic  surd. 

For,  in  simplest  form,  every  quadratic  surd  has  as  a  radicand  one 
or  more  prime  factors  raised  to  the  first  power  only. 

Two  dissimilar  surds  cannot  have  all  of  these  factors  alike,  and 
accordingly  their  product  must,  after  it  is  simplified,  have  at  least 
one  of  these  factors  to  the  first  degree  as  a  radicand. 

From  this  it  follows  that 

(iii.)  The  sum  or  the  difference  of  two  dissimilar  quadratic  surds 
cannot  be  equal  either  to  a  rational  number  or  to  a  single  su/rd. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


SURDS  443 

Va  ±  a/^  7^  c,     when    a  ^  h,  (1) 

and  Va  ±  v^  ^  V^,  a  7^  6.  (2) 

For  if  V^<  ±  V^  =  <?,     we  should  have 

a  ±  2Va  \/6  +  6  =  c^ 

it  -" 

Hence,  we  should  have  the  product  of  two  dissimilar  quadratic 
surds  equal  to  a  rational  number,  which  is  impossible  by  (ii.) 
above. 

It  follows  that  \fa  ±  ^/h  cannot  be  equal  to  c,  when  a  i^  b, 

A  similar  method  of  proof  holds  for  (2). 

(iv.)  The  square  root  of  a  rational  number  cannot  be  expressed  as 
the  sum  of  another  quadratic  surd  and  a  rational  number. 

That  is,  if  ^fa  and  ^/b  are  quadratic  surds,  and  c  is  any  rational 
number,  it  follows  that  a/S  t^  ^/b  +  c. 

For,  if  v^  =  a/^  +  c  (1) 

we  have,  squaring,         «  =  6  +  2  c^J'b  +  c\ 

Therefore,  v^  =  "^  ~  ^  ~ ""'  •  (2) 

u  c 

That  is,  if  (1)  be  true,  we  have  in  (2)  a  surd  number  V^  equal 
to  a  rational  expression,  which  is  impossible. 

Accordingly,  ^/a  cannot  be  equal  to  a/^  +  c. 

(v.)  In  any  equation  containing  quadratic  surds  and  rational 
numbers^  the  surd  numbers  in  one  member  are  equal  to  the  surd  num- 
bers in  the  other  member,  and  the  rational  numbers  in  one  member 
are  equal  to  the  rational  numbers  in  the  other  member. 

That  is,  if  v^  +  3/  =  v^  +  ^,  (1) 

it  follows  that  a;  =  «,  and  ?/  =  5,  where  a,  b,  x,  and  1/  are  all  com- 
mensurable numbers  and  a^^  and  Vx  are  surds. 

For,  ify:^b,\eti/=b±  n,  where  n  ^  0.  (2) 

Substituting  b  ±  /»  for  3/  in  (1),  we  obtain, 

^/x  +  b  ±  n  =  "s/a  +  b. 

Or,  ^/x  =  ^fa  =f  n^ 


444  FIRST  COURSE  IN  ALGEBRA 

which  is  impossible  by  Principle  (iv.)  above. 

Hence  we  cannot  assume  that  y  is  different  from  i,  as  in  (2) 
above. 

It  follows  that  y  =  h^  and  hence  a/«  =  \/«,  or  a;  =  a. 

(vi.)   If  N/^Tv^=a/^+ A/y,  (1) 

then  \a  —  ^h  =  v^  —  Vy,  (2) 

provided  that  a,  h,  Xy  and  y  are  commensurable,  and  that  a  >  V^. 

For,  from  (1),  a  +  a/^  =  ic  +  2v^  +  y. 

Hence,  by  Principle  (v.) 

a  =■  a;  +  y,  and  v^  =  2'^xy. 

Hence  a  —  Vf  =  jc  +  y  —  2v^. 

Or  Sla  —  ^/b  =  v^  —  Vy- 

51.*   Square  Root  of  a  Binomial  Quadratic  Surd  may  be 

obtained  by  applying  the  principles  of  §  50. 

Ex  1.   Find  the  square  root  of  14  +  2y^. 

Let                              Vl4  +  2  v^  =  ^/x+  ^/y.  (1) 

Then  by  (vi.),  ^14  -  2/y/33  =  y^  -  y^.  (2) 
Multiplying  the  members  of  (1)  by  the  corresponding  members  of  (2)  we 
have,                                  yi96  -  132  =  x  -  y. 

Or                                              X  -  2/  =  8.  (3) 

From  (1),  squaring,           14  +  2-^/33  =  a:  +  2^/xy  +  y.  (4) 

By  (v.),  a:  +  2/  =  14.  (5) 
Solving  (3)  and  (5),                           a:  =  11,     y  =  ^. 

Hence,  from  (1),        Vl4  +  2y'33  =  ^^  +  ^/^. 
52.*    Solution  by  Inspection. 

From  the  identity,  (v^  ±  Vbf  =  a  -{-h  ±  1^/ah, 

it  follows  that,  V«  ±  V^    =  \a-\-h±  "l^fah. 

It  should  be  observed  that  the  expression  a  •\-h  ±.  'i^ab  consists 
of  a  surd  term  ±  l^/ah  and  a  rational  binomial  a  •\-  h. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


SURDS  445 

The  radicand  of  the  surd  term  ±  2\/abis  the  product  of  two  fac- 
tors a  and  b  of  which  the  sum  is  the  rational  binomial  a  +  b. 

Hence,  to  find  by  inspection  the  square  root  of  a  binomial  surd 
which  is  a  square,  we  may  proceed  as  follows  : 

Transfoi^m  the  given  binomial  quadratic  surd  so  that  the  coefficient 
of  the  surd  term  shall  be  2 ;  then  find  by  inspection  two  factors  of  the 
radicand  of  the  surd  term  of  which  the  sum  is  the  rational  term  of  the 
transformed  binomial  surd. 

The  square  root  required  is  the  sum  or  the  difference  of  the  square 
roots  of  tJie  numbers  thus  obtained^  according  as  the  given  binomial 
surd  is  a  sum  or  a  difference. 

Ex.  2.   Find  by  inspection  the  square  root  of  53  —  IO-y/G. 

We  have  ^53  -  10//6  =  V^53  -  2 /y/ 150. 

The  two  factors  of  150,  of  which  the  sum  is  53,  are  50  and  3. 

Hence  V^53  -  lOy^  =  y^  -  ^-  5/^/2  -  /y/S. 

53.*  It  may  be  shown  that,  if  x  and  y  are  positive  rational 
numbers,  Six  ±  Vy  can  be  expressed  as  a  simple  binomial  surd, 
provided  that  ar^  —  y  is  the  square  of  a  rational  number. 

Exercise  XX.     15 

'  (This  exercise  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Find  the  square  root  of  each  of  the  following  binomial  quadratic 
surds: 

1.  8  —  2Vl^.  6.  19  -  Vl92.  11.  28  +  7a/12. 

2.  7  —  2a/I2-  7.  64+  6a/7.  12.  51  +  7a/8. 

3.  17  +  2^/70.  8.   18  -  8'\/5.  13.  88  -  9a/28. 

4.  5  +  \/24.  9.  32  +  10a/7.  14.  6  +  3a/3. 

5.  13  —  A/T68.  10.  18  -  3\/20.  15.  3  +  ^5. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


446  FIRST  COURSE  IN  ALGEBRA 


CHAPTER  XXI 

IMAGINARY  AND  COMPLEX  NUMBERS 

I.    Imaginary  Numbers 

1.  An  even  root  of  a  negative  number  cannot  be  expressed  either 
as  a  positive  or  as  a  negative  number.    (See  Chap.  XVIII.  §§  7,  14.) 

By  the  Law  of  Signs  in  multiplication  the  product  of  an  even 
number  of  positive  or  of  negative  numbers  is  positive;  hence  a 
negative  number  cannot  result  as  the  product  of  an  even  number 
of  positive  or  of  negative  foctors  alone. 

E.  g.  Since  (±  5)*  =  25,  -y/—  25  cannot  be  expressed  either  as  a  positive 
or  as  a  negative  number. 

2.  In  order  that  an  indicated  even  root  of  a  negative  number  may 
be  admitted  to  our  calculations,  we  shall  assume  that  the  identity 
iVaY  =  a  holds  without  exception  for  negative  as  well  as  for  posi- 
tive values  of  the  radicand  a.    (See  Chap.  XVIII.  §  10.) 

E.  g.  >y/^l  is  defined  to  be  such  a  number  that  its  square  shall  equal 
-  1,  that  is,  (V^n^)*  =  -  1. 

3.  An  even  root  of  a  negative  number  is  called  an  imaginary 
nuiuber, 

2nj 


E.  g.  V—  1)  V—  2>  V—  4,  V—  »>  and  y^—  a  are  all  imaginary 
numbers. 

4.  The  initial  letter  i  of  the  word  imaginary  is  commonly  used 
to  represent  a/— T,  which  is  taken  as  the  unit  of  imaginary 
numbers.     Hence  2^  =  —  1.     (See  §  2.) 

5.  Imaginary  numbers  have  no  existence  in  an  arithmetic  sense, 
and  hence,  when  first  introduced  into  mathematical  science,  were 
called  "imaginary"  numbers  before  their  meaning  and  use  in  con- 
nection mth  other  "  kinds  "  of  number  were  understood. 


IMAGINARY  NUMBERS  447 

6.  To  distinguish  them  from  imaginary  numbers,  all  other  num- 
bers previously  defined,  —  such  as  rational  or  irrational  numbers, 
whether  they  be  positive  or  negative,  integral  or  fractional,  —  are 
called  real  numbers. 

Certain  other  names  have  been  suggested  for  "  imaginary  "  num- 
bers and  "real"  numbers,  but  we  shall  employ  these  commonly 
accepted  terms. 

7.  In  order  to  be  able  to  operate  with  imaginary  numbers  by  the 
same  rules  as  with  real  numbers,  we  must  assume  that  all  positive 
or  negative  multiples,  or  fractional  parts  of  the  unit  of  imaginaries, 
V—  1  or  i,  are  numbers,  and  that  they  obey  all  of  the  Laws  of 
Algebra. 

E.  g.  Just  as 

4  =  1  +  1  -h  1  -I- 1,  so  4V~  =  ^^  +  V^  +  V^  +  V~; 
and  as 

I  =  i  +  i  +  i      so  |V=i:  =  iV=n[  + 1  v=^  +  iV=^- 

8.  Multiplication  by  i  may  be  defined  by  assuming 
that  the  Commutative  Law  holds  for  imaginary  num- 
bers : 

that  is,  a  X 

Or 

E.  g.    3  X  V^  =  V^^  X  3  =  V^+  V^^  +  V-^- 

By  multiplying  each  of  the  numbers  of  the  extended  _  ^. 

series  of  whole  numbers  by  i,  we  may  form  the  series  _  3^- 

of  purely  imaginary  whole  numbers.  -  4i 

9.  Powers  of  i.     It  should  be  observed  that 

— ooi 


-\-ooi 


"+  3i 

+      i 
±  Oi 


and  that  V^H"  V-  1  =  V{-  If  ^  ^+  1'  =  +  1. 

This  is  because,  whenever  the  radicand  is  known  to  be  the  square 
of  a  negative  number,  such  as  (—  1)^,  the  square  root  must  be  a 
negative  number,  —  1.     (See  Chapter  XVIII.  §  i:^).) 

We  may  obtain  the  following  powers  by  multiplication  : 


448 


FIRST  COURSE   IN   ALGEBRA 


w~i)' = (V^nv^)  =  (- 1)  v^i = -  v^, 

=  -  (V^iy  =  -  (- 1)  =  + 1, 

(V=rT)«  =  (V-  1)*  V-  1  =  a/=^. 

The  values  of  the  first  four  powers  of  V—  1  are  all  different,  and 
the  value  of  the  fifth  power  is  the  same  as  that  of  the  first.  Con- 
sequently, since  each  power  is  obtained  by  multiplying  the  power 
next  preceding  it  by  V—  1,  it  follows  that  the  results  obtained 
above  must  recur  in  groups  of  four  different  values. 

That  is, 


=  {*n+2  =  _^^ 


•   Z=  I 


;4n+3 


n  being  zero  or  any  positive  integer. 

From  the  reasoning  above,  it  follows  that : 

(a)  A?ii/  even  power  of  i  is  equal  to  one  of  the  real  numbers  —  1  or 
+  1,  and  any  odd  power  of  i  is  equal  to  one  of  the  imaginary  numbers 

+  V^^  m-  -  V^. 

(b)  According  as  the  remainder  obtained  by  dividing  the  exponent 
mof  a  given  power  of  i  by  4  is  1,  2,  3,  or  0,  the  value  of  i"^  is  i,  i^,  i^ 
or  i*  ;  that  is,  /,  —  1,  —  /,  or  +  1,  respectively. 

E.  g.  i^  =  i*  •  6+3  =  ?3  ^  _  i  .    iSi  =  H  •  8+2  =  ^-2  ^  _  1. 

10.  As  a  result  of  assuming  that  the  Commutative  Law  holds,  we 
have  the  Distributive  Law  {x  ±  y)  i  =  xi  ±  yi ;  and  also  the 
Associative  Law  xiyi  =  i\i/  =  —  xy. 

11.  Division  by  i.     To  conform  with  the  definition  of  division, 

—  must  be  such  a  number  that,  when  multiplied  by  i,  the  product 
is  ni» 


IMAGINARY  NUMBERS 


449 


By  our  definition,         n  X  i  =  ni. 
Hence  n  X  i  -r-  i  =  ni  ■ 


Or 


i 


12.  The  square  root  of  any  negative  number  is  an  imaginary 
number  and  may  be  expressed  in  the  form  V—  ci  =  i\/a. 

For,  since     (y^V^^Y  =  (Va)\V^iy  =  -a 
and  (V—  ciY  =  —  ^) 

it  follows  that         {V^^f  =  (V^V^^Y- 

Accordingly,  for  principal  values  of  the  roots, 

Of  the  two  values,  +  V—  «  and  —  V—  <^,  of  the  square  root  of 
any  given  negative  number  —  a,  the  first  one,  +  a/—  «,  is  selected 
as  denoting  the  principal  value  of  the  square  root  of  the  given 
negative  number. 

Ex.  1.         /v/-25  =  V^^C-  1)  =  ^25  V-^  =  5/v/^  =  5  i. 

Ex.  2.       3 V'^e  =  3>v/36(-  1)  =  3-v/36 V^  =  3  '  6-^/^  =  18  i. 

Mental  Exercise  XXL     1 

Reduce  each  of  the  following  powers  of  i  to  one  of  the  numbers 
1,  —  1,  i  or  —  i  : 


1.  P. 

2.  i'\ 

3.  i'\ 

4.  ^«^ 

5.  ^«^ 

6.  i'\ 


7.  P. 

8.  —  ^^ 

9.  -  2« 

10.  e". 

11.  z^'^. 

12.  ^l''^ 


13.  (V^Y- 

14.  (V^)^^ 

15.  (V-£)" 

16.  (V-  1)'' 

17.  (V^)'' 

18.  (V=^)'' 


19.  (V=^)'^ 

20.  (V^)". 

21.  -(.v/^)"'. 

22.  (V^y\ 

23.  (V=^)''. 

24.  (V=T)"l 


Express  each  of  the  following  imaginary  numbers  as  a  multiple 
of  the  unit  of  imaginary  numbers 


—  1 


25. 

V-  9. 

29. 

-3V-49. 

33.  4V-  121. 

37.  --3'\/-^'. 

26. 

V-  16. 

30. 

5V-  64. 

34.  V-rt^ 

38.  ^V-  /• 

27. 

V-  25. 

31. 

-7a/-81. 

35    \/—b'. 

39.  -aV-^;'. 

28.  2a/—  36.       32.  9V- 


100. 
29 


36.  2V- 


40.   V— 4«'^ 


450  FIRST  COURSE   IN  ALGEBRA 

41.  V-  49  b^  46.  a/^^.  51.   V—  9  bK  56.  b\/~Uz\ 

42.  aV-  x^.  47.   V^^.  52.   V-16c^  57.   {V^^f^ 

43.  yV^f,  48.   \/-rf^^  53.   V-  25</».  58.  (V^)^®. 

44.  -cV^c'^  49.  V-  ^".  54.  2V^^^U^\  59.  (V-c)^^ 

45.  V^^.  50.  V— 4  a*  55.  3V-49/.  60.  (V^/*. 

Addition  and  Subtraction  of  Imaginary  Numbers 

13.  Since  imaginary  numbers  may  be  written  so  as  to  appear  as 
arithmetic  multiples  of  the  units  /  and  —  /,  they  may  be  combined 
by  addition  or  subtraction  by  the  same  principles  as  real  numbers. 

In  all  operations  involving  imaginary  numbers,  it  will  be  conven- 
ient to  write  the  terms  of  given  expressions  as  multiples  of  the  unit 
of  imaginaries,  V—  1  =  '• 

It  should  be  observed  that  V--^a  =  V—  lV«  =  i^/a. 

=  bi         +  3  i  —  i  =7  i. 

\/-49//'^=4ai  -7bi=(4a-7 b)i, 
^  +  I'll  =  ^  +  (_  ^)  =  0. 
t's  +  i'-^  =  1  +  1  =  2. 

Exercise  XXI.    2 
Simplify  each  of  the  following  : 


1.  2V—  1  +  3 V-  1.  7.  2V-  81  +  3\/— I. 

2.  V^^  +  V^l.  8.  4V^  —  6V^n^. 


3.  V-  16  +  V-  9.  9.  V—  169  -  2V-  36. 

4.  V— 49  —  V—  25.  10.   V—  196  —  4\/^^- 

5.  V-64  +  V—  100.  11.  ttV—  100  -  2A/-49a^ 

6.  V—  121  —  V—  144.  12.  8V— 64ar'  —  10  a; V- 25. 

13.  V—  121  —  2^—49  +  3V^^. 

14.  13V^  -  V-  169/  +  6a/~  16/. 

15.  V—  144^2  —  V— 64  a*  —  V—  16  a«. 

16.  9V-  4  a'  -  4 aV-  9  «"«  -  a V-  36  a"". 

17.  V^^=^  -  V^  -  V==^. 


IMAGINARY  NUMBERS 


451 


18.  a/^^  +  a/—  18  —  V—  50. 

19.  V-  12  -h  5a/-  75  -  lOV- 


108. 


20.  3a/-  20  -  2 V-  45  +  5a/-  125. 

21.  ^«  +  e'  +  ^«  +  /^  28.  13  i^"-  -  31  P. 

22.  P  -  11  P.  29.  z^  +  e*  +  ^*. 

23.  3  ^»  -  4  ^^  +  5  i^  30.  ^^«  -  ^^*  -  ^l^ 

24.  6  —  i\  *  31.  8  ^'  -  4  i^  +  2  ^2. 

25.  7  *  —  ^^.  32.  ^*  +  z^  +  2^  +  «'• 


26. 
27. 


33.  P  - 

34.  2z2 


Multiplication  of  Imaginary  Numbers 

14.    In  case  either  the  multiplier  or  multiplicand,  or  both,  are 
imaginary  numbers,  the  following  principles  apply :  , 

(  (i.)  ^/7l^/^^=    VctiVb  =  iVab  =  V—ab, 

I  (ii.)     V—  aV—  b  =  iVa  i\/b  =  i^Vcib  =  ~Vab, 

Ex.1.         V^  X  V49    z=2ix7=l4i. 

Ex.  2.         V-2  X  V-^  =  *V2  X  t'v^  =  i'^^/lO  =  — v^lO. 


Exercise  XXL   3 
Simplify  each  of  the  following  : 

1.  2i  X  i.  6.  —  1  i  X  i. 

2.  3  /  X  i.  7.  4  ^  X  (—  3  i). 

3.  ^  X  6e.  8.  —  9  z  X  {—i). 

4.  4/  X  5e.  9.  —  10«  x(-2z). 

5.  8  /  X  3  i.  10.   V3  X  V^^. 

16.  —  V^^  X  V^^. 

17.  -\/^  X  (-  V^). 

18.  -  2a/^^  X  (-3\/^^). 

19.  2a/^^  X  3^/^^. 

20.  -V-^Te  X  V^. 

21.  V^^x  V-^- 


11.  V5  X  V^^. 

12.  V^^  X  V^. 

13.  V^^l  X  V^. 

14.  V^^  X  a/^. 

15.  V^  x  V^. 


22.  V^^  X  V—  10. 

23.  a/^  X  a/^^. 

24.  a/—  14  X  V^. 

25.  V^  X  a/^. 

26.  ia  X  i. 

27.  ih  X  i. 


452 


FIRST  COURSE  IN  ALGEBRA 


53.  V—  10  a  X  V—  '^  «. 

54.  V— 3^  X  V-6  6. 


55.  V-21<?X  V—  7  c?. 

56.  V—  iPy  X  V—  a% 


57.  ^/—xyz  X  V—  ais. 

58.  xV—yz  X  xyV^^. 

59.  xV^^  X  icV—  a*. 

60.  3  e  X  2  2  X  /. 

61.  5  e  X  3  ^  X  /. 

62.  6  e  X  4  2  X  /. 

63.  —  7  2  X  5  ^  X  ^. 

64.  —  8  e  X  3  e  X  2  /. 

65.  ia  X  ib  X  ic, 

66.  /a;  X  iy  X  iz. 

67.  2  /a;  X  4  ix  X  6  zaj. 

68.  ia^  X  ia^  X  ia. 

69.  V--^  X  V^^  X  V^. 

70.  \/^^  X  V^^  X  V^^. 

71.  a/- 2;")  X  V-49  X  V^^. 


V—  5  c  X  V—  6  c. 

78.  ^-^3^  X  V—^y  X  V—^z. 

79.  V—  3  «6c  X  ^/—2ab  X  V^^- 

80.  V=^^  X  V^'  X  \/^^. 

81.  (-\/:=^(-v=^(-V^^. 

82.  ia  X  ib  X  ic  X  id, 

83.  ib  X  ^c?  X  ix  X  iy. 

84.  2  2a  X  3  ih  X  im  X  2. 


72.  V-  64  X  V-  36  X  V-  9. 

73.  V—  12  X  V^  X  V^. 

74.  V—  14  X  V^  X  V^. 

75.  V-  15  X  V^  X  V^^. 

76.  V--^  X  V^^  X  V^^' 
11,  -V—'Zc  X  ^/-'6dX  V^^. 


IMAGINARY   NUMBERS 


453 


85.   iahc  X  iab  X  ia  X 


86. 

5  ^6  X  4  ib  X  ic  X  2. 

87. 

V-  2  X  V—  3  X  V-  4  X  V—  1. 

88. 

V-  5  X  V—  6  X  V—  2  X  V-  1. 

89. 

V—  6  X  V—  2  X  V-  3  X  V-  1. 

90. 

V—  10  X  V—  5  X  V—  2  X  V—  1. 

91. 

^*  X  i. 

97.  i^  X  i^  X  ^^ 

92. 

i'  X  i\ 

98.  —  ^«  X  ^^  X  /^ 

93. 

i'  X  i\ 

99.  3  z*  X  4  i^  X  z^. 

94. 

2i'X  4e». 

100.  bi^  X  4i*  X  3^^ 

95. 

-  7  i'  X  a 

/«.                            101.  -7e«  X  5?''  X  2 

96.  ^*  X  e«  X  ^^.  102.  z'  X  «*  X  ^«  X  i\ 

103.  2  2^  X  3  ^*  X  4  2«  X  5  i\ 

Division  of  Imagiuary  Numbers 

15.   In  case  either  the  dividend  or  divisor,  or  both,  are  imaginary 
numbers,  the  following  principles  apply  : 

V—  a  _  i^a  __  ..  la      .  I     a 

_  v« 


(i.) 
(ii.) 
(iii.) 


VI- v/- 


Ex.  1. 


Ex.  2. 


Ex.  3. 


^''^iV2. 


V 
J_ 

—  i 


V6  a/6 

^_*V2_,/2_1      /^ 
^  =  ^V^  =  V3  =  3^• 


-  1 


+  *. 


16.  In  the  last  example,  the  operation  of  "realizing"  the  im- 
aginary denominator,  or  making  it  "  real,"  suggests  the  operation 
of  "  rationalizing  "  an  irrational  denominator. 


454 


FIRST  COURSE   IN   ALGEBRA 


9V-  16  H-  I6V-  9. 


Exercise  XXI.    4 
Simplify  each  of  the  following : 

1.  V^^^^VTl.  20.  iV-  14 -f 

2.  V-39  H-  Vl3.  21. 

3.  V— 30  -r-  V3.  22. 

4.  V^^  ^  \/=^.  23. 

5.  V-  14  -7-  V^.  24. 

6.  V-  15  H-  V^.  25. 

7.  -  V-26  ^  V-  l.S.  26. 

8.  V— 28  -^  (-  V^^).  27. 

9.  V-20  -7-  \/=^.  28. 

10.  12e^3e.  29. 

11.  18eH-6/.  30. 

12.  24^-^8^.  31. 

13.  35  2  ^  5  i.  32. 

14.  V^  ^  V=^.  33. 

15.  V^^  -T-  V^^,  34. 

16.  V^a  ~-  \/=nO.  35. 

17.  a/- 11-f-  V^^.  36. 

18.  2\/^  -T-  \/^=^.  37. 

19.  SV^  -^  V^.  38. 


2. 


49  V—  25  -^  25  V—  49. 

V—  «6c  -r-  V—  a. 
— V  —  a;  -T-  V—  ^. 


39. 

2 

V-3 

40. 

4 

V-5 

41. 

6 

V-2 

42. 

12 

V-6 

43. 

11 

V- u 


44. 


45.  V 


46. 


47. 


48.  ^ 


-9 


V—  abc  -r-  V—  bed* 
a^—  a  -T-  hv—  h. 

iVi  -^  iVTb. 
W2  -7-  zVs. 
—  iVii  -^  «V3. 

49.  ^. 


50.  |. 


51.  |. 


11 


52. ; 


53   i? 

5^-  ,.10 


IMAGINARY   NUMBERS 


455 


55. 


56. 


17.  Both  negative  and  fractional  numbers  were  included  in  our 
extended  number  system  by  bringing  in  the  idea  of  measuring  dis- 
tances or  counting  in  opposite  directions. 

From  the  point  of  view  of  the  primary  numbers  1,  2,  3,  4,  etc., 
negative  numbers  and  fractional  numbers  both  have  an  existence  as 
imaginary  as  "imaginary  numbers." 

It  remains  for  us  to  show  that  imaginary  numbers  may  be  given 
a  graphical  interpretation. 

18.  By  means  of  the  principle  of  geometry  that  in  a  right  triangle 
the  square  on  the  hypotenuse  is  equivalent  to  the  sum  of 
tJie  squares  on  the  remaining  two  sides^  we  may  repre- 
sent graphically  any  surd  number. 

E.  g.  If  in  a  right  triangle  the  sides  including  the  right 
angle  are  each  one  unit  in  length,  the  hypotenuse  has  a 
length  represented  by  'y/2.     (See  Fig.  1.) 

Using  the  length  thus  found  for  v  2,  we  may  find  a  length  representing 
'V/a.     (See  Fig.  2.) 


-2  -V5  -yjl  -1 


^J2^fZ2 


Fig.  2. 


Fig.  8. 


By  setting  off  lengths  thus  found  along  a  straight  line,  as  in  Fig.  3, 
definite  points  can  be  located  on  the  line,  representing  +  a/2,  +  V3,  + 
^/1=  2,  -  V2,  -  Vs,  -  V4  =  -  2,  etc. 

By  separating  the  line  from  0  to  1  into  equal  parts,  we  may  locate  points 
representing  positive  fractions,  such  as  +  |,  +  i,  +  f ,  etc. 

In  a  similar  way  points  r(;presentiiig  negative  fractions,  such  as  —  ^,  —  f, 
^tc,  may  be  located. 


456  FIRST  COURSE   IN  ALGEBRA 

Graphical  Kepreseutation  of  Imaginary  Numbers 

19.  Consider  the  following  products  consisting  of  a  positive 
arithmetic  number  a  multiplied  by  —  1  and  by  (—  1)^. 

Multiplying  «  by  —  1,  once^  «  X  (—  1)  =  —  a. 

Multiplying  «  by  —  1,  twice,    «  x  (—  1)(—  1)  =  «  x  (+  1)  =  +  a. 

It  appears  that  multiplying  a  by  —  1  once  reverses  the  quality  of 
a,  while  multiplying  by  —  1  twice  successiveli/  reverses  and  then  re- 
stores the  quality  of  a. 

Accordingly,  if  +  «  represents  a  length  OAi,  measured  along  our 
"  carrier  line  "  in  the  ]X)sitice  direction,  since  multiplying  by  —  1 
reverses  the  quality  of  +  «,  to  represent  —  a  we  must  measure  an 
equal  length  OA3  in  the  opposite  or  negative  direction  from  the 
same  starting  point  or  origin,  0. 

We  may  think  of  the  line  a  as  having  made  a  "  half  revolution  " 
about  the  origin  0  from  the  position  0^1 1  to  the  position  OA^. 

Multiplying  —  a  by  —  1  will  pro-  _  ^  _^  ^ 

duce  a  second  "  half  revolution  ; "  ' ^1- 

hence  multiplying   -fr/Jby  —  1*  1 

twice  s^ivccestaively  may  be  thought  of  as  producing  a  "complete 
revolution  "  of  a  about  the  origin  0. 

20.  Observe  that  by  multiplying  -f  a  by  V—  1  =  /  once,  twice, 
three  times,  and  four  times  successively,  we  obtain  the  numbers  +  «», 

—  «,  —  ia  and  H-  a,  respectively,  as  follows  : 

ai  =  +  ia, 
aii  =—  a, 
aiii  —  —  ia, 
aim  =  +   a. 

It  appears  that  the  quality  of  a  may  be  reversed  either  by  mul- 
tiplying by  i  twice  or  by  multiplying  by  —  1  o?ice ;  and  the  quality 
of  a  remains  unaltered  when  a  is  multiplied  by  ifour  times  or  by 

—  1  twice  successively. 

If  a  represents  a  given  distance  measured  along  a  fixed  line  in 
one  direction,  it  may  be  seen  that  the  quality  of  a  may  be  reversed 
by  turning  a  about  one  of  its  end  points  through  a  half  revolution  ; 
and  the  quality  of  a  remains  unaltered  when  a  is  turned  about  one 
of  its  end  points  through  a  complete  revolution. 


IMAGINARY  NUMBERS 


457 


ia- 

V 

^3  -a    0 

V  —  ia 
J  A, 

Hence,  since  performing  the  same  algebraic  operation  two  times 
(multiplying  a  by  V—  1  two  times)  has  the  same  effect  on  a  as 
turning  a  about  one  of  its  end  points  through  a  half  revolution,  it 
may  be  seen  that  it  is  consistent  to  interpret  multiplying  a  by  i 
once  as  causing  a  quarter  revolution  of  the  line  a  about  one  of  its 
end  points. 

If  we  let  OAi  represent  the  original  posi- 
tion of  the  line  a,  then,  having  made  a  quarter 
revolution,  OA2,  will  represent  ia;  at  a  half 
revolution,  OAs  will  represent  —  a;  at  a  three- 
quarters  revolution,  OA^  will  represent  —  ia ; 
while  0^1 1  will  represent  the  position  of  the 
line  a  after  having  made  a  complete  revolution 
about  the  point  0. 

It  is  customary  to  suppose  the  line  OAi  to  swing  upward  and 
around  the  point  0  in  the  direction  opposite  to  the  movement  of 
the  hands  of  a  clock,  that  is,  counter-clockwise. 

It  follows  that  the  positive  imaginary  number  +  ia  may  be  repre- 
sented graphically  by  measuring  a  length  "  upward  "  along  a  line 

at  right  angles  to  a  in  its 
original  position,  OAi.  The 
negative  imaginary  number 
—  ia  will  accordingly  be 
represented  by  measuring 
downward  along  the  same 
perpendicular  line  a  dis- 
tance equal  to  a. 

The  line  AsOAi  is  called 
the  axis  of  real  numbers 
to  distinguish  it  from  the 
axis  of  imaginary  num- 
bers, A4OA2,  which  is 
drawn  at  right  angles  to  the 
line  A3OA1. 

We  may  represent  the  two  series  of  numbers,  "  real  and  imagi- 
nary," as  in  the  accompanying  figure.  It  should  be  observed  that 
the  series  of  real  numbers  and  the  series  of  imaginary  numbers 


\+coi 

f 

.+2i 


M 


-3  -2  -1 


■t.. 


+1    +2   +3 


+00 


-2z. 

I 
» 
I 

I 

Fig.  4. 


458  FIRST  COURSE  IN  ALGEBRA 

have  in  common  the  value  zero,  and  no  other  value.  This  value 
zero  is  represented  by  the  point  0,  which  is  the  intersection  of  the 
axis  of  real  numbers  and  the  axis  of  imaginary  numbers. 

21.  One  imaginary  number,  ia^  is  said  to  be  equal  to  another,  ib^ 
if  a  =z  I).  We  cannot  say  that  a  7-eul  number  is  equal  to,  greater 
than,  or  less  than  an  imaginary  number. 

22.  Negative,  fractional,  surd,  and  imaginary  "numbers"  are 
all  "  extended  "  or  "  invented  "  numbers  with  reference  to  arithme- 
tic numbers,  that  is,  to  "positive  whole  numbers"  which  alone  are 
the  result  of  counting.  Of  these  "  artificial  "  or  "  invented  "  num- 
bers, each  may  have  a  graphical  interpretation,  and  accordingly 
from  this  point  of  view  one  "  form  "  is  no  more  "  imaginary  "  than 
another. 

II.     Complex  Numbers 

23.  The  algebraic  sum  of  a  real  number  and  an  imaginary  number 
is  called  a  complex  number. 

E.  g.    2  +  y  ^,  5  -  8  i,  a  ±  ib. 

If  a  and  h  represent  real  numbers,  the  general  expression  for  a 
complex  number  is  a  -h  ib» 

In  particular,  if  a  =  0,  «  -f  ib  becomes  ib^  which  is  an  imaginary 
number. 

If  6  =  0,  «  -f  ib  becomes  «,  which  is  a  real  number. 

24.  Two  complex  numbers,  which  differ  only  in  the  signs  of  the 
terms  containing  the  imaginary  unit  ^,  are  called  conjugate  com- 
plex numbers. 

E.  g.    a  -{-ih,  and  a  —  ib  are  conjugate  complex  numbers. 

25.  Two  complex  numbers,  x  -\-  iy  and  a  +  ib,  are  said  to  be 
equal  if  their  real  terms,  x  and  a,  are  equal,  and  their  imaginary 
terms,  iy  and  ib,  are  equal. 

.^  26.  In  order  that  a  complex  number,  x  -f  iy^  shall  be  equal  to 
zero,  it  is  necessary  and  sufficient  that  the  real  part  shall  equal  zero 
and  the  imaginary  part  shall  equal  zero. 

For  if  a;  -f  /y  .-^  0  =  0  4-  /O,  then  by  §  25,  £c  =  0,  and  also  y  =  0. 

27.  If  either  aj  or  y  is  infinite,  the  complex  number  x  -f-  iy  is  said 
to  be  infinite. 


COMPLEX   NUMBERS  459 

28.  The  Four  Fuudamental  Operations  Involving  Com- 
plex Numbers  are  defined  by  assuming  that  the  fundamental 
laws  of  algebra,  as  proved  for  real  numbers,  apply  also  for  complex 
numbers. 

29.  Addition  and  Subtraction  of  Complex  Numbers.  The 
sum  of  two  complex  numhers  is  in  general  a  complex  number  obtained 
by  adding  the  real  parts  and  the  imaginary  parts  separately. 

For,  since  complex  numbers  are  assumed  to  obey  the  Fundamental 
Laws  of  Algebra,  it  may  be  seen  that 

(a  +  ib)  ±  (x  +  iy)  =  (a  ±x)  -{-  i(b  ±  y). 

The  principle  applies  for  three  or  more  complex  numbers. 

Ex.  1.       (3  +  5^=1:)  +  (6  -  2^^  =  (3  +  6)  +  (5  -  ^)^/^^ 

=  9  +  3  *. 

30.  The  sum  of  two  crnijugote  complex  numbei's  is  a  real  number. 
For,         {a  +  ib)  +  {a  —  ib)  =  {a  +  a)  +  i{b  -b)  =  2a. 

31.  Multiplication  of  Complex  Numbers  is  defined  by  as- 
suming that  the  Distributive  Law  for  Multiplication  (Chapter  V. 
§  21)  applies  to  complex  numbers. 

(a  +  ib)(x  +  iy)  =  (a  +  ib)x  +  (a  +  ib)iy 
=.ax  +  ibx  +  aiy  +  ibiy 
=  (ax  —  by)  +  i(bx  +  ay). 

32.  It  may  be  shown,  by  applying  the  Associative  Law  (Chapter 
III.  §  4),  that  in  general  the  product  of  two  or  more  complex  numbers 
can  be  expressed  as  a  complex  number. 

Ex.  2.    (4  +  7  0(2  -  5  0  =  8  +  14^  -  20i;  -  35  i^  =  43  -  6*. 
Ex.  3.    (a  +  ihy  =  a^  +  2  ai6  +  %%'^  =  a^  -b'^+2  iah. 

33.  The  product  of  two  conjugate  complex  numbers  is  a  number 
which  is  real  and  positive. 

For,  {a  +  ib)(a  -  ib)  =  a^  -  i'b^  =  a'  +  P. 

Ex.  4.    (-5  +  2a/=^)(-5-2V^)-^(-5)2-22(V^)'=25+12=37. 

34.  Division  of  Complex  Numbers.  The  quotient  obtained 
by  dividing  one  complex  number  by  another  can  be  expressed  as  a 
comj)lex  number. 


460  FIRST  COURSE   IN   ALGEBRA 

For, 


a  +  ib  __(a  +  ib)(x  —  ii/) 
X  +  iy~  (x  -\-  iy){x  —  iy) 

_  {ax  +  by)  4-  i{bx  —  ay) 
~  x'  +  y'^ 


ax  ■\-  by       .bx  —  ay 

35.  From  the  reasoning  above  it  appears  that  a  fraction  the 
denominator  of  which  is  a  complex  number  can  be  expressed  as  a 
complex  number. 

2-3/_(2-30(4-5i) 
•    •  4  +  5t~(4  +  5i)(4-6i) 

_-7-22t 
~         41 

Ex.  6.  7  -  2V--6  ^         (7  -  2^V6)(VlO  +  3zV5) 

_  7 yio  +  6  V30  +  ^(21  a/5  -  4//I5) 
~  55 

Exercise  XXI.    5 
Simplify  each  of  the  following  expressions  : 

1.  (2  +  30  +  (5  +  2  e).  6.  (9  -  5  2)  -  (5  +  9 1). 

2.  (5  +  80  +  (2  +  40-  7.  (3  +  4a/^)  +  (7+2V=^). 

3.  (7  -  10  0  +  (2  +  9  0-  8.  (8-7\/^)+(8-10'v/^). 

4.  (4  -  11  0  -  (6  +  3  0-  9.  (4-12a/=4)-(9-11  V=4). 

5.  (1  -  12  0  -  (12  -e).  10.  (6-2v^)-(5  +  4V^). 

11.  (2  +  V^)  +  (4  +  V=T6). 

12.  (3  +  V^=^)  +  (6  +  V-  4). 

13.  (5  -  V-36)  -  (6  +  V-  49). 

14.  (25  -  ^-36)  -  (36  -  V-  25). 

15.  (8  -  a/^)  -  (18  -  V-  18). 

16.  (5  +  2a/^  +  (7  -  3\/-  12). 

17.  (11  +  2\/-  20)  -  (4  -  3V-  180). 

18.  (6  -  5  V-  28)  -  (12  +  W-  63). 


COMPLEX   NUMBERS 


461 


(a  +  3  0  +  (2  +  bi), 
(a;-4^•)-(2/+6^). 
(m  -  8  i)  -(n-10  i). 
(2a  +  5bi)  +  (Qa  —  bi). 
(5  +  Si)i6-^i). 
(-6  +  7  0(-  6  -  7  ^). 
(7  +  4^)(7-4^). 

(8  +  V^^)(8  -  V^^ 

(7  +  3V=^)(7  -  3a/-  5). 

(3  +  2*)(4  +  5i). 

(8-6^)(2  +  5  0. 

(2-^Xl-2^). 

(V2  +  2')(^;^  +  0-  

(\/2  +  V=n")(V2  -  V-jO- 

(V3_+\/=r5)(V3- V-5). 

(3V5  +  2V3)(3a/5  -  V3).  

(6V7  +  8\/^(6V7  -  SV-  y). 
(2  +  3  0(V5  4-«V6). 

(1  +  zy. 

(1  +  i)\ 
(3  +  4zy. 
l-^(2  +  3z). 

5  --  (2  -y^^)^_ 

10  -T-  (V5  -  V-  2). 
51  -=-  (1  -  5  V^^)- 
2  + Si 
2  — Si' 

9+  V^ 
8+  y/^ 
1  +  V-  1 . 

1-  v^ 

48.  (/^  +  aV'^^)(«  +  bV^^)'    

49.  (t/V^  +  x\/^^)Q^V^  —  3/V-  ^)- 


19. 

20. 

21. 

22. 

23. 

24. 

25. 

26. 

27. 

28. 

29. 

30. 

31. 

32. 

33. 

84. 

35. 

36. 

37. 

38. 

39. 

40. 

41. 

42. 

43. 

44. 

45. 
46. 
47. 


462  FIRST  COURSE  IN  ALGEBRA 

50.  (-  1  4-  V^)(--l  -  V^. 

51.  (aV^^i  -  bV^'- 

52.  (1-/7(1  +  0- 

53.  (1  -  2)(2  -  0(3  -  e). 

54.  (1  +  ^Xl  -3f)(l  +  5e«). 

Complex  Factors  of  Rational  Integral  Expressions. 

36.  The  method  of  Chap.  XX.  §  48,  may  be  applied  to  obtain 
factors  of  expressions  of  the  form  ax^  +  bx  +  c  which  are  the  prod- 
ucts of  complex  factors. 

The  following  identity  was  obtained  in  Chapter  XX.  §  48 : 

L        2a  2a       JL        2a  2a       J 

If  the  expression  ^^  —  4  ac  be  negative,  Vb"  —  ^ac  will  be  imagi- 
nary, and  accordingly  the  factors  of  ax^  ■\-  bx  -\-  Cy  represented  by 
the  expressions  in  square  brackets,  will  be  complex. 

Ex.  1.   Factor  a;^  +  3z  -|-  4. 

We  may  complete  the  square  with  reference  to  x^  +  3  a;  by  using  3  a:  as  a 
finder  term  as  follows : 

w    u  3a:      3 

We  have  t;—  =  t:  • 

2a;      2 

Hence,  a;^  +  3a; -h  4  =  x^  +  3ar  +  (|)2  _  |  +  4, 

=  (^  +  1)'  +  ^,    _ 
=  (x  +  |)2_(y_7)2, 

=  (^  +  f+^i^)(^4-|-^). 
Square  Root  of  a  Complex  Number. 

37.*  Corresponding  to  the  principle  employed  when  finding  the 
square  root  of  a  simple  binomial  surd  (see  Chap.  XX.  §  50  (vi.)),  we 
have  the  following 

Principle :  If  the  square  root  of  a  complex  number  can  be  expressed 
as  a  complex  number ,  then  the  square  root  of  the  conjugate  complex 
number  can  also  be  expressed  as  a  complex  number. 

That  is,  if  VcT+Tb  =  Vx -\-  iVy,  (l) 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


COMPLEX   NUMBERS  463 

it  follows  that  V«  —  2^  =  V^c  —  Wy-  (2) 

If  a,  ^,  a;,  and  ^  are  real  numbers,  we  obtain,  by  squaring  both 

members  of  (1),  a-\-ih  =  x  —  y-\-^  ia/xij. 

Hence,  by  §  25,  ^  =  a;  —  ?/,  and  also  h  =  2^/xy. 
Accordingly  we  may  construct  the  expression 
a  —  ib  =  X  —  y  —  2  i\/xy 
=  {y/x  —  i^/yf. 
Hence,  V«  —  if>  —  Vx  —  lA/y- 

38.*   It  follows  that  the  square  root  of  a  complex  number  can  be 
expressed  as  a  complex  number. 


For,  from  §  37,  if  ya  +  lb  =  Vic  +  Wy^  (1) 

it  follows  that  V«  —  ib  —  V-^  —  Wy,  (2) 

in  which  a.,  b,  x,  and  y  are  real  numbers. 

Hence,  from  (1)  and  (2)  we  obtain  by  multiplication, 


V  («  +  ib)(a  —  ib)  =  (yx  +  iVy)(Vx  —  Wy\ 
or,  V«'  +  b''  =  x^-y,  (3) 

Squaring  both  members  of  equation  (1), 

a-^{b  =  x  —  y+2  iVxy.  (4) 

Hence,  by  §  25,  a  =x  —  y.  (5) 

Solving  equations  (3)  and  (5)  for  x  and  ?/,  we  have 

x  =  - 2 (6)  and  7/  = ^ (7) 

Accordingly,  from  (1), 

V«  +  lb  =  V ^ +  ^V ^ (8) 

Since  a  and  ^  are  real  numbers,  it  follows  that  a^  +  //  is  positive ; 
hence  a/oM-^  is  a  real  number,  and  accordingly  the  right  member 
of  (8)  is  a  complex  number. 

Ex.  2.   Express  aJ—  i  as  a  complex  number. 
We  may  write  V  —  i  =  y'O  —  i. 

*  This  eection  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


464  FIRST  COURSE   IN  ALGEBRA 


Comparing  \/0  -  i  with  the  form  ^a  +  bi,  it  appears  that  a  =  0  and 
6  =  —  1 ;  hence,  carrying  out  the  process  shown  above,  or  substituting  im- 
mediately in  (8),  we  obtain, 

_(t-i)Va 

■"  2 

Exercise  XXI.     6 
Find  complex  factors  for  each  of  the  following  : 

1.  a'  +  ^-.  4.  aj2-4«-8. 

2.  25ic2+  1.  5.  '■2x'+  3aj  +  4. 

3.  2!^  +  6a;+  10.  6.  9a;'  — 8a;—  7. 

(The  foUowing  examples  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Express  as  complex  numbers  the  square  roots  of  the  following 
complex  numbers  : 


7.  -  1  +  2V-  2.  10.  —  36  -  OV-  48. 

8.3  —  4/.      11.  —  I  —  6\/— 10. 

9.  41  —  14  V-  8.  12.-1  —  2eV2. 

Graphical  Representation  of  Complex  Numbers. 

39.  Applying  the  method  of  §§  19,  20,  for  the  representation  of 
an  imaginary  number,  we  may  represent  a  complex  number  a  +  ib 
graphically  by  means  of  a  point,  B.  This  point  is  located  by  first 
measuring  a  distance  OA  equal  to  a  from  the  origin  0  along  the 
axis  of  real  numbers^  and  then  from  the  point  A  thus  reached 


COMPLEX  NUMBERS 


465 


{a+ib) 


AX 


Fig.  5. 


measuring  a  length  AB,  equal  to  b,  in  a 
direction  parallel  to  the  axis  of  imaginai-y 
numbers^  that  is,  at  right  angles  to  the  axis 
of  real  numbers.     (See  Fig.  5.)  — 

The  point  B  may  be  called  the  graph 
of  tUe  complex  number  a  +  ib.  It  may 
be  seen  that  the  graphs  of  all  complex  num- 
bers which  have  equal  real  parts  a,  lie  on  the  same  straight  line,  A  B, 
parallel  to  the  axis  of  imaginary  numbers,  OY.  It  may  also  be 
seen  that  the  graphs  of  all  complex  numbers  which  have  the  same 
imaginary  part,  ih,  lie  on  the  same  straight  line,  passing  through  the 
point  B,  parallel  to  the  axis  of  real  numbers,  OX.     (See  Fig.  6.) 


{a'-^ib')^    {a-\-ib')^ 

I 

I 

u 


\X 
I 

[a-ib")^ 


Fig.  6. 


40.  *  The  length  of  the  line  OB  may  be  shown  by  principles  of 
elementary  geometry  to  be  equal  to  the  positive  value  of  V^^  +  ^"^. 
The  positive  value  of  V^^  +  b'^  is  called  the  modulus  of  the  com- 
plex number  a  +  ib.     (See  Fig.  7.) 

E.  g.  The  modulus  of  eitlier  of  the  conjugate  complex  numbers,  A  +  ^i 
<  ,r  4-3  i,  is  +  y/4^  +  3^  =l  +  5. 

41.='^  The  modulus  of  a  complex  number  may  be  taken  as  repre- 
senting the  absolute  or  arithmetic  value  of  the  given  complex 
number. 

42.  *  One  complex  number,  a  +  ib,  is  said  to  be  numerically 
equal  to,  greater  than,  or  less  than  another  complex  number,  x  +  iy, 
according  as  the  modulus  a/«^  +  b^  of  the  first  is  equal  to,  greater 
than,  or  less  than  the  modulus  V^c^  +  if  of  the  second. 

*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 
j]0 


466 


FIRST  COURSE   IN  ALGEBRA 


43.*  It  may  be  shown  that  the  points  representing  all  complex 
numbers  having  equal  moduli  lie  on  the  circumference  of  the  same 
circle,  while  all  those  representing  complex  numbers  of  greater  or 
less  absolute  value  lie  without  or  within  the  circle  according  as  their 
moduli  are  greater  than,  or  less  than,  the  modulus  of  the  given 
complex  number. 

E.  g.    The  complex  numbers  4  +  3  i,  4  —  3  z,  —  4  +  3  i,  —  4  —  3 1,  3  +  4  i, 
3-4i, - 
+  5. 


3  +  4i,  —  3  —  4t,  etc.,  all  have  tlie  same  modulua,  ^^4*  +  3^  = 
Each  of  these  complex  numbers  may  be  represented  by  a  definite 
point  situatetl  on  a  circle,  of  which  the  center  is  the  origin  and  the  radius 


is  5.     This  circle  passes  also  through  the  points  representing 


the  real  num- 
(See  Fig.  8.) 


bers  +  5  and  —  5  and  the  imaginary  numbers  +  5 1  and  —  5  i. 

All  numbers  of  greater  or  less  absolute  value  may  be  represented 
by  points  situated  outside  of,  or  inside  of  this  circle,  respectively. 

E.  g.    The  point  representing  the  complex  number  6  +  2 1,  of  which  the 
modulus  is  y'4()  =  6  +,  lies  without  the  circle,  while  the  point  representing 

the  complex  number  4  -\-  2i,  of 
which  the  modulus  is  'v/20  =  4+, 
lies  within  the  circle.    (See  Fig.  8.) 

44.  By  applying  the  Principle 
of  No  Exception,  we  have  ex- 
tended our  idea  of  number  from 
the  primary  arithmetic  whole 
number,  to  negative,  fractional, 
irrational,  imaginary,  and  finally 
to  complex  number. 

We  have  shown  in  §§  29-35 
that  the  fundamental  operations 
of  addition,  subtraction,  multi- 
plication, and  division,  when 
applied  to  complex  numbers,  result  in  general  in  complex  numbers. 
It  may  be  shown  by  principles  and  methods  beyond  the  limits  of 
elementary  algebra  that,  whenever  the  direct  operations  of  addition, 
multiplication,  and  involution,  or  the  indirect  operations  of  subtrac- 
tion, division,  and  root  extraction,  are  applied  to  any  of  the  kinds  of 


Fig.  8. 


*  This  section  may  be  omitted  when  the  chapter  is  read  for  the  first  time. 


REVIEW  467 

number,  including  complex  number,  the  results  in  each  case  lead  to 
no  new  kind  of  number. 

The  number  represented  by  {a  +  iby'^'^  is  a  complex  number. 

Accordingly,  the  complex  number  may  be  regarded  as  the  most 
general  kind  of  tmmOer,  and  with  its  inclusion  the  number  system 
of  algebra  is  complete. 

Mental  Exercise  XXI.     7     Review 

1.  Show  that  a;®  +  6  A-^  +  9  a;*  is  the  square  of  a  binomial. 

2.  Find  the  continued  product  of  x^  —  i/%  x^  -f  7/\  x^  +  ?/*  and 
««  +  f. 

Obtain  each  of  the  following  quotients  : 


3.  {a 

-b)- 

-.{b-a). 

5.  (5  - 

-^- 

(^-5). 

4.  (c- 

-4)- 

-  (4  -  c). 

6.   (x'- 

-f)- 

^(ij-  x). 

Square 

each  of  the  following  : 

7.  a. 

.12.  -  i. 

17.  .06. 

22.  .01. 

8.  b\ 

13-  §• 

18.  ai 

23.  c~K 

9.  c\ 

14.  .2. 

19.  bl 

24.  «f"l 

10.  i. 

15.  .3. 

20.  c\ 

25.  -  3. 

11.  h 

16.  .5. 

21.  (#. 

26.  -  x^ 

Find  the  value  of 

-?■ 

-  (i: 

2 

29. 

2 

3^* 

Express  as  a  single  power  of  5  : 

30.  25^     125' ;     625*. 
Find  the  value  of 

31.  3'  —  2^  32.  2'  +  2-^ 

Express  as  a  power  of  a  base: 

33.  (2^^.  34.  (y^y.  35.  (z^. 

36.  Find  the  values  of  {^y  and  -  2'^ 

Express  the  following  with  positive  exponents  : 

37.  a-^  -T-  6"l  38.  x'"-  4-  y-^.  39.   (1/m)-''  -^  (1  /n)-\ 


468  FIRST  COURSE  IN  ALGEBRA 


Show  that  the  following  identities  are  true  : 

42.  {x  -  y){z  -  y){x  ^z)  =  {x-  y){y  -  z){z  -  x). 
Simplify  each  of  the  following  : 

43.  \^HM\  44.   v^l6^.  45.   '^27  twV. 

46.  Regarding  x  as  the  unknown,  solve  -  H !--=:</. 

XXX 

Distinguish  between 

47.  fa  and  ai         48.  x'^  and  —  x. 

49.(f)-l.(i)-,a„d(f)(-l).     52.2a„dl. 
-  a         a^ 

50.  rt-l  +  />-!  and -4-7-  no     ^      r:     r:0    nS    ^         ^  ^ 

a  +  6  53.  0  •  5,  5^  0^  -  and  -. 

()  5 

51.  —  1  -^  a  and  1  -f-  a~\  54.  (—  a)~^  and  —  a~\ 

55.  (-  ^)-«  and  -  b-\ 
Which  of  the  following  complex  numbers  have  equal  moduli  1 
56.  3  +  4  ?,  4  —  3  /,  5  +  12  /,  —  3  +  4  /,  —  12  —  5  i. 
Simplify  each  of  the  following  : 

58. 
59. 
60. 
61. 


62. 


i+  1 

i 
i  +  3 

4  +  i 

i 
i  +  i 

20 
a 
ft-" 

63. 

y 
y 

z 

y 

•  w 

w 

64. 

z 

~' 

65. 

(«-■ 

f  «')(«='  - 

a-^. 

66. 

(f 

"Xf 

-,). 

67. 

(a^- 

-?,f)^ 

.QUADRATIC  EQUATIONS  469 


CHAPTER  XXII 

EQUATIONS  OF  THE  SECOND  DEGREE  CONTAINING 
ONE  UNKNOWN 

1.  If  the  members  of  an  equation  containing  the  second  power  of 
one  unknown  quantity,  «,  are  rational  and  integral  with  respect  to  x, 
and  if  they  contain  no  powers  of  x  other  than  the  second  and  the 
first,  the  equation  is  said  to  be  of  the  second  degree  with  reference 
to  ic,  or  quadratic. 

E.  g.  a:2  4-  5  a:  +  6  =  0, 

a;24- 3ar  =  7 -2a:2  +  5a;. 

2.  From  every  quadratic  equation  containing  one  unknown  quan- 
tity, X,  may  be  obtained,  by  applying  the  principles  of  Chapter  X,  an 
equivalent  equation  of  the  standard  form  ax^  ■\-  bx  -\-  c  =  0. 

In  this  equation,  which  will  be  referred  to  as  the  Standard 
Quadratic  Equajtion,  a,  6,  and  c  represent  known  quantities,  a 
being  positive  and  different  from  zero.  The  first  term,  ax^^  repre- 
sents the  sum  of  all  of  the  terms  containing  oi? ;  the  second  term, 
bx^  is  the  sum  of  all  of  the  terms  containing  x  to  the  first  power,  and 
the  third  or  known  term,  c,  stands  for  the  sum  of  all  of  the  terms 
that  are  free  from  the  unknown,  x, 

E.  g.  Froin  4a;  —  2 a;* +  8  =  3  —  2a;  —  5  x^  may  be  derived  the  equiva- 
lent quadratic  equation  in  standard  form,  3a;2  4-6a:-|-5=:0. 

In  the  equation  3  a;*^  -|-  6  a;  -f-  5  =  0,  the  numbers  3,  6,  and  5  take  the  place 
of  a,  6,  and  c  respectively  in  the  standard  quadratic  equation  ax^  _(-  fex  -f-  c  =  0. 

3.  In  the  standard  quadratic  equation  am?  +  bx  +  c  =^  0,  either 
of  the  letters  b  ox  c  may  represent  zero,  causing  the  corresponding 
terms  to  disappear  from  the  equation,  but  a  cannot  be  zero^  for  in 
that  case  there  would  be  no  term  containing  y? ;  hence  the  equation 
would  not  be  quadratic. 


470  FIRST  COURSE  IN  ALGEBRA 

In  all  that  follows  we  shall  therefore  assume  that  a  7^  0,  and  that 
a,  6,  and  c  are  all  real  quantities. 

4.  If,  when  reduced  to  the  standard  form  rt'ar  +  bx  +  c  =  0,  & 
quadratic  equation  contains  all  of  the  terms  represented  by  ax^,  hx 
and  r,  it  is  said  to  be  complete. 

If  either  of  the  terms  represented  by  bx  or  c  be  missing,  the  equa- 
tion is  said  to  be  incomplete. 

E.  g.  7a:*  +  3 a?  —  5  =  0  and  x*  —  3a:4-10  =  0are  complete  quadratic 
equations,  while  3i*  — 4  =  0  and  9x^  =  22;  are  incomplete  quadratic 
equations. 

Graphs  of  Quadratic  Equations. 

5.  The  graph  of  a  quadratic  eciuation  ha\nng  the  form  of  the 
standard  quadratic  equation  ax^  -\-  hx  -\-  c  =  0  may  be  obtained  as 
follows : 

Let  y  represent  the  value  of  the  expression  aa^  +  bx  +  c  when  a 
particular  number  is  substituted,  for  x.  The  value  assigned  to  x 
and  the  corresponding  value  calculated  for  y  may  be  taken  as  the 
x-coordinate  and  the  y-coordinate  respectively  of  a  point  on  the 
graph  of  the  given  quadratic  equation  a.i^  +  bx  +  c  =  y.  By  as- 
signing different  values  to  x  and  calculating  corresponding  values 
for  yy  the  coordinates  of  as  many  points  on  the  graph  of  the  given 
quadratic  equation  as  may  be  required  may  be  obtained. 

E.  g.  We  may  obtain  a  portion  of  the  graph  ol  the  tpuulratic  equation 
x*  +  4  X  =  5  a.s  follows : 

Transposing  the  terms  of  the  given  equation  to  the  first  member  and 
representing  the  value  of  this  first  member  by  t/,  we  have  x^  -\-  4x  —  5  =  y. 

Substituting  different  values  for  x  in  the  equation  x2  +  4x  —  5  =  2/we 
may  calculate  corresponding  values  for  y. 

If            X  =  0,  we  have  0  +     0  —  5  =  y.  Hence,   —  5  =  y. 

x=l,  1+4-5  =  i/.                       0  =  y. 

x-%  2^  +  4-2  -  5  =  y.                       7  =  1/. 

X  =  3,  32  +  4.3  _  5  =  y.                     16  =  y. 

etc.  etc.                                  etc. 

The  values  shown  in  the  accompanying  table  may  be  readily 
calculated. 


QUADRATIC  EQUATIONS 


471 


x2+4^ 

-  5  =  ,/ 

X 

y 

6 

66 

5 

40 

4 

27 

3 

16 

2 

7 

1 

0 

0 

-5 

-1 

-8 

-2 

-9 

-3 

-8 

-4 

-5 

-5 

0 

-6 

7 

(- 


5£) 


m 


Fig.  1.    a:2  +  4x-5=y. 


The  general  shape  of  the  graph  may  be  determined  as  follows  : 

By  factoring  the  first  member  of  x^  +  4:X  —  b  =  y^ 
we  obtain  (a;  +  5)  (a;  —  1)  =^y. 

For  all  values  of  x  greater  than  +  1  or  less  than  —  5,  the  two  factors 
(x  -\-  5)  and  (a:  —  1)  have  like  signs,  and  consequently  the  corresponding 
values  of  y  are  positive,  and  the  points  on  the  graph  of  which  these  values 
of  X  and  y  are  coordinates  must  lie  above  the  axis  of  X. 

For  all  values  of  x  lying  between  +  1  and  —  5,  the  factors  (x  +  5)  and 
(a;  —  1)  have  opposite  signs,  one  positive  and  the  other  negative.  It  follows 
that  the  product  {x  +  b){x  —  1),  represented  by  y,  is  negative.  Hence  the 
values  of  i/,  obtained  by  substituting  values  lying  between  +  1  and  —  5  for 
X,  must  be  negative,  and  the  points  on  the  graph  of  which  these  values  of 
X  and  y  are  coordinates  must  be  found  helow  the  axis  of  X. 

The  graph  crosses  the  axis  of  X  in  the  two  points  (—  5,  0)  and  (+  1,  0), 
and  in  no  others,  as  shown  in  Fig.  1. 

The  values  —  5  and  +  1,  which  locate  the  points  of  "crossing,"  are  the 
values  which  reduce  the  expression  x^  +  4  x  —  5  to  zero,  and  hence  are  the 
roots  of  the  equation  x^  +  4  x  —  5  =  0. 

•  6.  We  have  illustrated  in  this  example  the  principle  that  every 
quadratic  equatimi  containing  one  unknown  has  two  roots,  and  can- 
not have  more  than  two  roots. 


472 


FIRST  COURSE   IN  ALGEBRA 


7.  Approximate  values  for  the  real  roots  of  a  quadratic 
equation  may  be  obtained  by  first  constructing  its  graph,  and  then 
measuring  the  distances  along  the  axis  of  A"  from  the  origin  to  the 
intersections  (if  there  be  any)  of  the  graph  and  this  axis,  and 
estimating  the  numerical  values  of  x  in  terms  of  the  scale  unit 
according  to  which  the  graph  is  constructed. 

8.  If,  instead  of  crossing  the  axis  of  X,  the  graph  lies  wholly 
upon  one  side  of  it,  and  simply  touches  it  in  one  point,  it  is  con- 
venient to  say  that  there  are  two  values  of  x  which  satisfy  the 
equation,  but  that  these  values  are  equal. 

E.  g.  The  solutions  of  a:*  +  4  x  +  4  =  0,  which 
is  equivalent  to  (x  +  2)(aj  +  2)  =  0,  are  x  =  —  2, 
and  X  =  —  2. 

By  referring  to  Fig.  2,  it  will  be  seen  that  the 
graph  of  x* -1-4x4-4  =  1/  touches  the  axis  of  X  at 
the  point  (—  2,  0),  and  does  not  cross  it. 

9.  If  the  graph  neither  crosses  nor  touches 
the  axis  of  X,  there  are  no  "  real "  points  of 
intersection  with  the  axis,  and  in  such  cases 
it  will  be  found  that  there  are  no  "real" 
solutions  to  the  equation,  but  there  are  two 
"  imaginary "  values  which  may  be  found 
by  solving  the  equation. 


Fig. 2.   x'^  +  ^x  +  A  =  y. 


E.  g.    By  methods  which  will  be  shown  later, 
the  solutions  of  x^  +  4  x  +  5  =  0  are  found  to  be 

X  =  -  2  +  V"-^  ^^^^  X  =  -  2  — y/^. 
These  solutions  are  complex  numbers,  and  it  will 
be  seen,  by  referring  to  Fig.  3,  that  the  correspond- 
ing graph  of  x'-^  +  4  X  +  5  =  2/  neither  crosses  nor 
touches  the  axis  of  X  ;  that  is,  the  points  of  crossing 
are  "imaginary." 


I.   Incomplete  Quadratic  Equations 

Fig.  3.    x^  +  '^x  +  b—y, 
10.   Any  incomplete  quadratic  equation  in  which  the  first  power 
of  the  unknown  is  missing,  may  be  reduced  to  the  form 


QUADRATIC  EQUATIONS  473 

ax''  +  c  =  0.  (1) 

We  have  assumed  a  to  be  different  from  zero  (see  §  3) ;  hence, 
dividing  both  members  by  a  and  transposing,  we  may  write 

.^  =  ^.  (2) 

In  this  form,  the  general  solution  of  the  equation  may  be 
obtained  by  either  of  the  following  methods. 

11.   First  Method.     We  may  employ  the  principles  of  fac- 

toring  by  first  expressing  as  the  square  of  its  square  root, 


at  is. 

?Kv/v)' 

Hence,  we  have 

-(v/~y- 

(3) 

Transposing, 

'■-(v'-7=»- 

(4) 

Accordingly,     fx  +  y -^)  (x  -  \-^)  =  0.  (5) 

The  single  equation  in  this  form  is  equivalent  to  the  two  sepa- 
rate equations  formed  by  equating  to  zero  each  of  the  factors. 
(Compare  with  Chap.  XII.  §  48.) 

Hence  we  may  write 

x+)J—^  =  0,  and    x-\/^^  =  0.  (6) 

The  solutions  of  these  equations  are  found  to  be 


=-v^. 


and  a;  =  +l/  — .         (7) 


It  should  be  observed  that  the  two  roots  are  equal  in  absolute 
value,  but  opposite  in  sign. 

If  c  and  a  have  unlike  signs,  both  roots  will  be  real,  but  if  c  and  a 
have  like  signs,  both  roots  will  be  imaginary. 


474  FIRST  COURSE   IN   ALGEBRA 

12.    Second  Method.    By  extracting  the  square  roots  of 

—  c 
both  members  of  x^  =  —  (see  (2)  §10),  we  may  obtain 

Or 

V    a 

which  is  a  convenient  abbreviation  for  the  separate  equations, 


▼     a  y    a 


(See  (7)  §  II.) 


Observe  that,  while  it  is  not  incorrect  to  write  the  double  sign  ± 
before  both  members  of  the  equation  after  extracting  the  square 
roots,  it  is  unnecessary. 

This  is  because  the  equation  x=  ±y  —  might  have  been  written 

±  x=  ±  y  —  >  which  is  a  convenient  abbreviation  for  the foUow- 
▼    a 

ing  set  of  four  equations  : 

-x=  +  \/^-    (3)  —  =  -sl^-    (4) 

By  changing  the  signs  of  the  terms  in  both  members  of  (3)  and 
(4),  we  obtain  (2)  and  (1)  respectively. 

Hence,  when  extracting  the  square  roots  of  both  members  of  an 
equation,  it  is  sufficient  to  write  the  double  sign  ±  before  the  square 
root  of  one  member  only. 

13.  Observe  that,  since  there  can  be  two,  and  only  two  square 
roots  of  a  given  quantity,  an  incomplete  quadratic  equation  of  the 
type  an?  +  c  =  0  can  have  two,  and  only  two  roots. 

Ex.  1.    Solve  the  incomplete  quadratic  equation, 

5a:2-50=x2a:2  +  25. 

Transposing,  collecting  terms,  and  dividing  the  terms  of  the  resulting 
members  by  the  coefficient  of  x^,  we  obtain  the  equivalent  equation 

x2  =  25. 


(-5.0) 


QUADKATIC  EQUATIONS 

Extracting  the  square  roots  of  both  members, 

X  =  ±  b. 
Accordingly  the  solutions  are 
a:  =  +  5,     and     x  =  -  5.     (See  Fig.  4.) 

By  substitution,  these  values  are  found  to 
be  solutions  of  the  original  equation. 

Substituting  +  5, 

5(+  5)2  _  50  =  2-52  +  25 
75  =  75. 
Substituting  —  5, 

5(-  5)2  -  50  =  2(-  5)2  +  25 
75  =:  75. 
Ex.  2.    Solve     3  a;2  =  (3  +  .r)(3  -  x). 
We  may  derive,     3  x^  =  d  —  x'^ 
Hence,  x^  =  |. 

Accordingly,  x  =  ±  ^. 

The  given  equation  is  found  to  be  satisfied 
by  both  of  these  values. 

Substituting  +  f ,  Substituting  -  |, 

mr  =  ('^  +  l)(3  -  f )  3(-  ly  =  (3  -  |)(3 


475 


Fig.  4.     x-  —  26  z=  y. 


I) 


V=i 


Ex.  3.    Solve  a;2  +  1  =  0. 

Transposing,  x^  =  —  1. 

Extracting  the  square  roots,     x  =  ±  ^/^^. 

The  solutions,  x  =  -\-  y"—  1  and  x  =  —  /y/^T 
(see  Fig.  5),  are  both  found  to  satisfy  the  original 
equation. 


-V=i 


Fig.  5.     x^ -\-  I  =  y. 


Substituting  +  /y/—  1, 
(V=^)'  +  1  r.  0 

-1  +  1  ::=0 
0  =  0. 


Substituting  —  y—l, 
(-  a/^)'  +1  =  0 
-1+1=0 
0  =  0. 


14.   Any  incomplete  quadratic  equation  containing  no  term  free 
from  X  may  be  reduced  to  the  form 

ax'  +  bx  =  0.  (1) 


476  FIRST  COURSE  IN  ALGEBRA 

In  this  form  its  solution  may  be  obtained  by  factoring. 

Factoring  (1),  we  have  x  {ax  +  ^)  =  0,  (2) 

which  is  equivalent  to  the  set  of  two  separate  equations, 
iP  =  0,         and        ax  -\-  b  =  0. 

The  solutions  of  these  equations  are  found  to  be 


12.0) 


and 

X  = 

d 

(3) 

Ex.  4.  Solve 

3x2  = 

6x. 

Transposing  and  factoring, 

3x(x-2)  =  0. 

Hence,        a:  =  0, 

and 

a:  -  2  =  0. 

Hence  the  solutions 

are 

a:  =  0, 

and 

x  =  2. 

(See  Fig.  6.) 

These  values  are  found  by 

substitution  to  i 

satisfy 

the  given  equation. 

Substituting  0, 

Substituting  2, 

0  =  0. 

3  •  22  =  6  •  2 
12=  12. 

Fig.  6.    3xi-Qx  =  y. 


15.  By  the  Principles  of  Equivalence,  roots  have  neither  been 
gained  nor  lost  in  making  the  different  transformations  in  the  pre- 
ceding examples ;  hence  the  solutions  shown  are  the  only  ones  which 
satisfy  the  given  equations. 

Exercise  XXII.     1 

Solve  the  following  Incomplete  Quadratic  Equations,  verifying 
results  : 

1.  Gar' -54  =  0.  11.  Ux"  +  1  =  6x^  +  43. 

2.  ISar'  =  64  -  icl  12.  6  ar'  -  5  =  ar^  +  45. 

3.  3ar*-  16  =  a;2+  16.  2ar'- 7_ 

4.  5ar'-8=:2a;2+  19.  13       ~ 

5.  ear'- 8  =  12a;''— 11.  3ar'+  13  _ 

6.  3a;='+  11  =  7a!2-5.  16        ~" 

7.  5x^-1  =  7x^-9.  , .    a!^  +  ,5       a;'  +  2 

lo. 


8.  (x+  5y=  lOa^-f  34.  2  3 

9.  (a;+3)^-6(a;+3)  =  7.         _    3x^-2   .   x'' +  4 
10.  ax'-b  =  a-  hx\ 


16.  "-^^        +  '^—^  =  4. 


QUADRATIC  EQUATIONS 


477 


17.  2(x  -  l)(2i«  +  1)  +  (8  +  a!)(10  -x)=Ax^+  29. 

18.  2(x  —  S)(x  +  4)  =  (a;  +  2)(x  +  5)  -x^  —  6x, 

19.  (x  +  2)(ar^  +  4)  ==  (a;  +  1)'  +  a;  -  2. 

20.  bx^={a-  b)\a  +  b)  -  ax\ 


II.     CoxMPLETE  Quadratic  Equations 

16.  After  having  transformed  a  given  quadratic  equation  into  an 
equivalent  quadratic  equation  in  standard  form,  ax^  +  6ic  +  <?  =  0, 
the  solution  may  be  obtained  by  different  methods.  These  methods, 
however,  are  all  based  upon  "  factoring." 

17.  If  the  factors  of  the  expression  represented  by  ax^  -\-  bx  +  c 
can  be  readily  obtained  by  inspection,  the  method  shown  in  Chap. 
XII.  §  47,  may  be  applied  to  the  equation  ax^  +  6ic  +  c  =  0,  and  we 
can  obtain  its 

Solution  l)y  Factoring 

Ex.  1.    Solve  a;2  +  4  a:  -  5  =  0. 

Factoring,  (a:  +  5)(a;  —  1)  =:  0. 

This  single  eqnation  is  equivalent  to  the  set  of  two  linear  equations 
obtained  by  writing  the  factors  separately  equal  to  zero. 

That  is,  X  +  5  =  0,  and         a;  -  1  =  0. 

From  which  a;  =  —  5,      and  x  =  +  1. 

(Conq)are  with  §  5  ;  see  also  Fig.  1.) 

Ex.2.    Solve   6x2  +  3=  lla:.  y| 

In  standard  form,     6  ic^  —  1 1  x  -|-  ,3  =  0. 
Factoring,  (3 «  -  1)(2 a;  -  3)  =  0. 

This  efjuation  is  equivalent  to  the  set  of  two 
linear  equations, 

3  X  -  1  =  0;     and     2  x  -  3  =  0. 
The  solutions  are 

X  =  \,     and  X  =  |. 

(See  Fig.  7.) 

Verifying  these  results  by  substituting  in  the 
given  equation,  we  have,  ^'^-  ^-   6  r  —  llx  +  3 

Substituting  \,  Substituting  |, 

6(^)2  +  3=  n(^)  6(1)2  +  3^11(1) 

¥  =  ¥•  ¥  =  ¥• 


(>f.O) 


478 


FIRST  COURSE  IN   ALGEBRA 


Exercise  XXIL    2 

Solve  the  following  Complete  Quadratic  Equations  by  the  Method 
of  Factoring,  verifying  results  : 


1.  x''-x  =  20. 


13.  3aj2  +  a;-  14  =  0. 


2. 

iB^  +  20  =  9aj. 

14. 

Sar*- 3a;- 2  =  0. 

3. 

a;2_4«  =  45. 

15. 

6a;^+  7aj=5. 

4. 

x'Jt  3ar  =  40. 

16. 

12  a;'' +  10  =  23  a;. 

5. 

15a;  =  ic^  +  36. 

17. 

8a;''  =  9(3a;-  1). 

6. 

110  =  0-^ -a-. 

18. 

(a;  +  3)-'  +(x-  4)'  =  65. 

7. 

ic(a?—  10)  =  11. 

19. 

(a;+  2)^-_17  =  2(a;-l)2. 

8. 

x{x  -  12)  =  -  27. 

20. 

(a;-3)'-3(a;-5)^-4  =  0. 

9. 

xix  +  15)  =  54. 

21. 

x^  -^^  ax  —  X  =  a. 

10. 

x(x-  14)  =  51. 

22. 

x"^  +  a  =  a^  +  X, 

11. 

2x'-{-  5a;  +  3  =  0. 

23. 

x^  +  x(h  —  a)  =  ah. 

12. 

3a-''—  7  a;  +  2  =  0. 

24. 

^x  +  a;^  =  c^x  +  &. 

18.  If  the  factors  of  the  expression  represented  by  ay?  +  bx  ■]-  c^ 
which  is  the  first  member  of  the  standard  quadratic  eciuation 
ax'^  -b  bx  +  c  =  Of  cannot  be  readily  obtained  by  inspection,  we 
may  obtain  the 

Solution  by  Completing  tbe  Square 

19.  The  expression  "  completing  the  square  "  may  be  given  a 
geometric  significance. 

By  attaching  a  strip  a  units  in  width  to  each  of  two  adjacent 
sides  of  a  square,  of  which  each  side  is  x  units  in  length,  a  geometric 
figure  may  be  constructed  which  is  made  up  of 
parts  represented  by  x^,  ax,  and  ax,  that  is, 
x^  +  2  ax,  as  shown  by  the  heavy  lines  in 
Fig.  8. 

The  figure  maybe  made  a  "complete"  square, 
each  side  of  which  is  (x  +  a)  units  in  length, 
by  adding  the  part  represented  by  a^. 

By   adding   a'^  to  a;^  +  2  ax  we   obtain   the 
algebraic   trinomial  x^  +  2  ax  +  a%   which    is 
the  square  of  the  binomial  x  +  a. 

The    algebraic    operation    of   obtaining    the    trinomial    square 


ax         a 

a2l 

X 

X^            X 

ax 

Fig.  8. 


QTTADRATIC  EQUATIONS  479 

Qi?  +  2ax  -h  a%  by  adding  the  square  a^  to  the  binomial  x^  +  2  ax, 
is  commonly  called  "  completing  the  square  "  with  reference  to  the 
binomial  x^  +  2  ax. 

Ex.  1.    Complete  the  square  with  respect  to  the  binomial  x^  -\-  6x. 

The  term,  the  square  of  which  must  be  added  to  the  binomial  x^  -\-  (5x  to 
obtain  a  trinomial  square  of  the  form  x^-\-6x-\-{  y,  may  be  found  by 
dividing  6  x  by  twice  the  square  root  of  the  term  x^. 

That  is,  1^  =  3. 

Adding  the  square  of  3  to  the  binomial  x^  -^  6  x,  we  obtain  the  required 
trinomial  square  a:^  +  6  a:  +  3^  =  x^  +  6  a:  +  9. 

Ex.  2.    Complete  the  square  with  respect  to  the  binomial  25  a^  +  40  ab. 

We  have  - — --  =4  6. 

2  •  5a 

Accordingly  the  required  trinomial  square  is 

25  a2  +  40ab  +  (46)2  =  25a^  +  40  a6  +  1662. 

Mental  Exercise  XXII.   3 

Complete  the  square  with  respect  to  each  of  the  following  bi- 
nomials : 

17.  25  z^-  GOzw. 

18.  64  a'' +  4.Sab. 

19.  9a^+  66  «6. 

20.  121  (r^  —  Ucd. 

21.  xY  —  2  xy. 

22.  c2^  +  2  cc?. 

23.  2r)/P_4Q^^^ 

24.  81  6-2^/2 -36  cd. 


1. 

a;^  +  2x. 

2. 

xf  +  47/. 

3. 

z'^-^z. 

4. 

a"  -\-  \2a. 

5. 

W-  Ub, 

6. 

c"—  16  c. 

7. 

^2+  18  c?. 

8. 

/-20^. 

9. 

8U'-  18  A. 

10. 

121^2 -22  c, 

11. 

h''  +  2  he. 

12. 

x^  +  22  xy. 

13. 

a^-2^ah. 

14. 

4aj^—  12a^. 

15. 

9A'+  60  M. 

16. 

4  ^/^  +  28  hy. 

25. 

49  ^V+  70  ^c. 

26. 

144  cV  +  120  ex. 

27. 

16  6^  +  8  h\ 

28. 

9  c*- 12c^c?2. 

29. 

4W*  +  36A^F. 

30. 

36  c«  -  12  cW 

31. 

25/^8+  30^*)?:*. 

32.  49  d^  +  84 


480  FIRST  COURSE  IN  ALGEBRA 


33.  81a^'"4-36a5~. 

35. 

c^y^H-  2c«y^. 

34.  lOOaj^V-60 

a^y".             36. 

121«*y"+44ar^y. 

37.  a^  +  3a% 

38.  z^-lz. 

49.a^  +  |. 

56.^       'f- 
5 

39.  y'+5y. 

40.  a^  +  a. 

50.  3^  + |. 

57.  4a;2  +  r 

41.  4a^+  5a. 

42.  9  6^  +  8  6. 

43.  16r»—  IOp. 

52.  z.^+\^. 

53.  a'  +  *5"• 

58.  9y«  +  |- 

59.  16c^+  f- 

5 

60.  4«,^       I'". 

0 

44.  25cr*-5rf. 

45.  36  A^-  13^. 

46.  9  m^  +  m. 

47.  16  w**-  w. 

-^'-*-•\^ 

61.  9a^+*'^ 

48.  a^  +  |. 

55.  <.-^^ 

62.256^      2f 

20.  To  solve  a  complete  quadratic  equation  in  the  standard  form 
ax-  +  l/x  +  c  =  0,  hy  the  method  of  "  completing  the  square,"  our 
problem  is  to  obtain  an  equivalent  equation,  one  member  of  which 
contains  all  of  the  terms  in  which  x  appears,  and  which  has  the  form 
of  a  trinomial  square. 

The  values  for  x  may  then  be  found  by  the  methods  used  in  the 
previous  sections  of  this  chapter. 

Ex.  1.    Solve  a:^ -I- 6a;  =  7. 

To  have  the  form  of  a  trinomial  square,  the  first  member  must  contain 
two  terms  which  are  squares  and  positive,  and  another  term  which  is  twice 
the  product  of  the  square  roots  of  these  two. 

The  first  term,  a;^,  is  the  square  of  x.  Hence,  using  the  remaining 
term,  6x,  as  a  "finder  term"  (see  Chap.  XII,  §  21),  we  may  obtain  the 
sciuare  root  of  the  missing  "  square-term  "  by  dividing  the  "  finder  term  " 
by  twice  the  square  root  of  the  first  term. 

That  is,  1^  =  3. 

By  adding  the  square  of  3  to  a;^  +  6  a:,  we  obtain  the  complete  trinomial 
square  x^  +  6  a:  +  3^. 


QUADRATIC  EQUATIONS 


481 


We  may  obtain  an  equation  which  is  equivalent  to  the  one  given,  having 
this  expression  as  a  first  member,  by  adding  to  both  members  of  the  given 
equation  3'-^  or  9. 

That  is,  a:2  +  6  X  +  32  =  7  +  9. 

Or,  (a: +  3)2^=42. 

Extracting  the  square  roots  of  both  members,  we  obtain 

x-\-3  =  ±4, 

which  is  a  convenient  abbreviation  for  the  set  of  two  equations, 

a;  +  3  =  +  4,     and     a:  +  3  =  -  4. 
The  solutions  are 

X  =  -\-  I,     and  X  z=  —  7. 

These  values  are  found  by  substitution  to  be  solutions  of  the  original 
equation. 

This  equation  may  also  be  readily  solved  by  factoring.  The  student 
should  obtain  the  graph. 


Ex.  2.    Solve 


a:2_8ar  +  ll  =0. 


The  first  term,  x%  is  a  square  and  is  positive  ;  hence,  using  the  second 
term  —  8a;  as  a  "  finder  term,"  we  may  obtain  the  "missing  term  "  which 
is  required  to  complete  the  s([uare,  with  reference  to  the  binomial  x^  —  8  a;. 

That  is,  =^  =  -  4. 
Hx 

Adding  the  square  of  —  4  to  both 

members,  and  transposing  +11  to  the 

second  member,  we  obtain  the  equiva- 

ent  equation. 


a:2_8a;+16  =  -ll  +  16. 
Hence,        (x  —  4)^  =  5. 
Extracting  the  square  roots, 
x-4:  =  ±  /y/B. 

It  follows  that  a;  —  4  =  +  y^,  and 
a;  -  4  =  -  /y/5. 
Hence, 


(4+V5,0) 


Fig.  9.    x^-Sx+n=y. 


x  =  4-\-  /y/5,     and     a:  =  4  -  ^5.     (See  Fig.  9.) 
These  solutions  may  be  verified  by  substituting  in  the  given  equation. 

Substituting  4  +  y^.  Substituting  4  —  a/5, 

(4  +  v^)2  -  8(4  +  /y/5)  +  11=0        (4  -  ^5)2  -  8(4  -  ^/6)  +  11=0 

21  +  8/y/5  -  21  -  8/v/5  =  0  21  -  8v^5  -  21  +  8/v/5  =  0 

0  =  0.  0  =  0. 

31 


482 


FIRST  COURSE   IN   ALGEBRA 


Although  thp  values  x  =  4  +  /y/5  and  x  =  4  —  /y/5  are  mathematically 
exadf  it  is  often  convenient  to  use  approximate  results  that  are  correct  to 
some  specified  number  of  decimal  places. 

Finding  the  value  of  ^5  correct  to  four  places  of  decimals,  approximor 
tions  to  the  true  values  of  x  obtained  above  are 

a-  r=  4  +  2.2360  +  and  x  =  4  -  2.2360  +, 

that  is,     x  =  6.2360 +,  and  a;  =1.7640+. 

Ex.3.     Solve  8x2- 7a:  -  1  =0. 

Observe  that  the  equation  as  written  contains  no  term  which  is  "  en- 
tirely "  a  square. 

By  either  dividing  or  multiplying  both  members  by  such  a  number  as 
will  transform  the  term  Sx^  into  a  square,  an  equivalent  equation  may  be 
obtained  which  can  be  solved  by  the  method  of  completing  the  square. 

Dividing  8  x^  by  either  8  or  2,  we  obtain  the  squares  x^  or  4  x^y  respec- 
tively, or  by  multiplying  by  either  2  or  8,  we  obtain  the  squares  IGa:^  or 
64  x\  respectively. 

We  may  obtain  an  equation  containing  the  term  x^,  equivalent  to  the 
given  equation,  by  dividing  the  separate  terms  by  8. 

Accordingly,  transposing  the  known  term  to  the  second  member,  we 

7^_  1 
8  ~8* 
Y  I  To  find  the  term  required  to  "  complete  the 

7  X 

we  may  use —  as  a  "  finder  term," 

o 


obtain 


i-W 


square 

iliO)   as  follows : 
_7x 

8 


2x  =  - 


7x 


-  —  +  (- 

8   ^^ 


S-2x-      ^^' 
Completing  the  square   with  reference  to 

x^  — — ,  by  adding  to  both  members  (—  ^^y  = 

+  2^>  ^'^  obtain  the  equivalent  equation 

Or,  (a:_J^)2  =  ^8^. 

Extracting  the  square  roots  of  both  mem- 
bers, we  obtain, 

l-y  Hence,     x  -  ^^  =  +  j\,     and    x  -  J-^  =  -  ^. 
Accordingly,     x  =  I,  and  x  =  —  ^. 

(See  Fig.  10.) 
These  values  are  found,  by  substitution,  to  satisfy  the  given  equation. 


Fig.  10.   8  x2  -  7  ar 


QUADRATIC  EQUATIONS 


483 


Ex.  4.    Solve  x^+x  +  l=0. 

X 

Usiiifj  a:  as  a  "  finder  term,"  we  have  --—  = 
°  2a: 

Adding  tlie  square  of  \  to  both  members 
of  the  equation,  and  transposing  the  known 
term  to  the  right  member,  we  obtain, 

x^  +  x  +  {^y  =  -  1  +  i, 
which  may  be  written  in  the  form 

Extracting  the  square  roots  of  both  mem- 
bers, we  obtain, 

a:  +  4  =  ±  V^- 
Hence, 

a;  +  4  =  +  \/^,    and   x-\-\  =  -  ^^/^. 

Hence  the  required  solutions  are 

a;  =  -  4  +  4  V^   and  x=-^-  ^\/^. 

(See  Fig.  11.) 


->^+KV=3 


'H-Vif^ 


Fig.  11.    x^  +  x+l=y. 


These  values  will  be  found,  by  substitution,  to  be  solutions  of  the  original 
equation. 

Substituting  ~  2  +  4  V~"  ^> 

(-4  +  4  V=^)'  +  (-  4  +  4  V^)  +  1  =  0 

i  -  4  \/^  -  f  -  4  +  4  V^  +  1  =  0 

0  =  0. 

Similarly,  substituting  —  4  —  4  V—  3,    we  obtain     0  =  0. 
Exercise  XXII.    4 

Complete  the  square,  and  solve  each  of  the  following  equations, 
verifying  all  results  obtained.  "Whenever  surds  appear  in  the 
exact  solutions,  approximate  values  correct  to  four  places  of  decimals 
should  be  obtained. 

1.  x^  +  2x=  15.  8.  a;2  —  12  =  4a;. 


2.  ar'  +  4aj  =  60. 

3.  x^  —  Qx=  16. 

4.  x^  +  8x  =  65. 

5.  cc^  — 2a;  =  8. 

6.  x""  +  10a;  =  11. 

7.  a;^  -  6  a;  =  —  8. 


9.  x^+  1  =  2x. 

10.  a;2  — 51  =  14a;. 

11.  x^—  242  =  11a;. 

12.  x{x  +  6)  =  —  9. 

13.  x(x—  12)  =  64. 

14.  63- 2a;  =  a;2. 


484  FIRST  COURSE  IN  ALGEBRA 

15.  x^+lx+  5  =  0.  24.  9a;'  +  GiB  -  4  =  0. 

16.  JB^  +  6a;  +  10  =  0.  25.  Gaj*^  +  a;  =  6. 

17.  7(x  -  3)  =  x^  26.  5a;'  -  a;  =  i. 

18.  (a;-2)(a;-3)  =  4. 

20.  3  af  +  a;  =  4.  2 

21.  5ar*  — 6a;+  7  =0. 

22.  6ar»+ 10a;  =  9.  o«^2,16iK_. 

23.  12a;'  -  13a;  =  32.  28.  a;  +  —  -  4. 

The  methods  employed  in  the  solution  of  numerical  equations 
may  be  applied  also  to  literal  equations. 

Ex.  29.   Solve  ax"  -  (a  -  b)x  -  b  =  0. 

We  may  obtain  an  equivalent  equation  containing  the  square  aH^  by 
multiplying  every  term  of  the  given  equation  hy  a. 
Hence,  we  have,  a^  —  a(a  —  b)x  —  ab  =  0. 

Using  —  a(a  —  h)x  as  a  "  finder  term,"  we  may  obtain  the  term  required 
to  complete  the  square,  as  follows: 

—  a(a  —  b)x  _  —(a  —  b) 
2ax        ~        2 
Completing  the  square  with  reference  to  a^x^  —  a{a  —  b)x,  and  transpos- 
ing the  known  term  —  ab  to  the  second  member,  we  obtain, 

aV  -  aia  -  b)x  +  (-  ^)'  =  +  <.6  +  (-  ^^T-^. 

Or,  aV-a(a-b):c  +  (^^  =  +  ab+(^^. 

Writing  the  first  member  as  the  square  of  a  binomial  and  combining  the 
terms  in  the  second  member,  we  obtain 

-by      4 ah  +  (a  -by 


(         a-  b\' 
Or,  (ax-^y  = 


4 
(a  +  by 
4 


-^                                                         a  —  ba  +  b 
Hence,  ax —  =  ±  —^  ' 

This  is  a  convenient  abbreviation  for  the  set  of  two  separate  equations, 

a  —  h          a->rb             .                 a  —  b  a-^h 

«^--2-  =  +  -Y-'      and      ax ^  =  --^. 


QUADRATIC  EQUATIONS  485 

We  obtain  the  solutions  as  follows  : 

«  —  fta  +  6  a  ~h      a  4-b 

a  —  b  -\-  a  -{-  b                                      a  —  b  —  a  —  b 
ax  = ax  = — 


2a  -  2b 

ax  =  -^  ax  =  — —— 

^  2 

b 

x  =  I.  X  — 

a 

These  values  will  be  found,  by  substitution,  to  satisfy  the  given  equation. 

30.  ar^  -  2  c?ic  =  c^  -  d^,  34.  «V  -  2  «a;  =  6  —  a^. 

31.  a;^  —  4  ca;  =  12  c".  35.  11  Ma;  =  6  ic^  -  7  m". 

32.  ar*  -  3  a^  =  2aa;.  36.  4(3 a;^  -  5 d'')  =■  dx. 

33.  x^+^ax  +  V'  =  0.  37.  2(6a;'  +  5  A^^  +  23  hx  =  0. 

38.    i^-(a  +  h)x  +  bab  =  ^ia"  +  ^'). 

21.  Hindu  Method.  In  the  method  employed  for  solving  the 
preceding  equations  it  will  be  observed  that,  to  complete  the  square, 
it  was  necessary  to  add  the  square  of  a  fraction  whenever  the  quo- 
tient obtained  by  dividing  the  "  finder  term  "  by  twice  the  square 
root  of  the  term  containing  x^  was  fractional. 

Quadratic  equations  may  be  solved  by  a  method  employed  by 
the  Hindus  which  allows  of  the  completion  of  the  square  without 
introducing  fractions  during  the  process. 

The  solution  of  the  following  equation  illustrates  the  Hindu 
Method. 

Ex.1.    Solve  3a;2_  13  a; +11  =  0. 

The  first  terra  may  he  made  a  square,  36  a;^,  by  multiplying  by  four 
times  the  coefficient  of  a;^,  that  is,  by  4  x  3  =  12. 

Accordingly,  multiplying  all  of  the  terms  of  the  given  equation  by  12, 
we  obtain  the  equivalent  equation 

36a:2-  156a:  +  132  =  0. 

We  can  find  the  number,  the  square  of  which  must  be  added  to  complete 
the  square  with  reference  to  the  binomial  36  a;^  —  156  a;,  as  follows : 

-  156  a; 
2-6ar 

It  will  be  observed  that  —  13  is  the  coefficient  of  x  in  the  given  equation. 
Adding  the  square  of  —  13  to  both  members  of  36  a:^  —  156  a:  +  132  =  0 
and  transposing  132  to  the  second  member,  we  obtain 


486  FIRST  COURSE  IN  ALGEBRA 

36x2  -  156  X  +  169  =  - 132  +  169. 
Hence,  (6  a:  -  13)2  =  37. 

Therefore,  6ar  -  13  =  ±  ^^7. 

Hence,        6  x  -  13  =  +  ^^7,       and       6  x  -  13  =  -  \/37« 

The  exact  solutions  of  these  equations  are 

13  +  V37  1  13  -  V37 

X  = ^ —     and  X  = . 

6  6 

By  extracting  \/37»  correct  to  four  places  of  decimals,  we  obtain  the 

following  approximate  values: 

13  +  6.0827  +  ,  13  -  6.0827  + 

X  = and        X  = ~ 

6  6 

Or,  x=:  3.1804 +,  and  x  =  1.1528 +. 

The  student  should  verify  the  exact  results  by  substituting  in  the  given 
equation. 

22.  It  should  be  observed  that,  when  the  Hindu  Method  is  applied 
to  the  solution  of  a  quadratic  equation  in  thefm^m  ax^  +  ^a;  +  c  =  0, 
the  separate  terms  are  all  multiplied  by  four  times  the  coefficient  a 
of  x^y  and  the  square  is  completed  with  reference  to  ay?  +  hx  by 
adding  the  square  of  the  coefficient  of  x^  represented  by  b. 

Exercise  XXII.    5 

Solve  each  of  the  following  equations  by  the  Hindu  Method, 
verifying  all  exact  rational  results.  Whenever  surds  appear  in  the 
eocact  solutions,  approximate  values  correct  to  four  places  of  decimals 
should  be  obtained. 

1.  3ic2+  10a +  8=0.  9.  (2  a; -7)2  =  6a;. 

2.  8a;^  +  26a;  +  15  =  0.  10.  (a;  +  l)(2a;  +  ^)=x^  -W. 

3.  5ar^- 16a;+  121  =0.  11.  3(a;+ l)(a;— l)  =  2(27--5a;). 

4.  Sar' +  50  =  25a;.  12.  (2a;+ l)(a;+ 2)  =  (a;- 1)1 

5.  15ar*+  16a;  =15.  13.  —-\  =  ^ 


x^       X      20 

6.  10ar^-9  =  43a;.  14.  Y'l'^Yl 

2  1 

7.  25ar'+  20a;  =  33.  ^^-11+  ^ 

8.  a;-l=20a^.  le.  ^^^^^  +  i  =  ? . 


20       4       5 

c^_^_2( 
3        7"~21 

11  "^  11~5 

«-5) 

30  6 


QUADRATIC  EQUATIONS  487 

General  Solution  of  aa?^  +  bx  +  c  =  O 

23.    The  general  solution   of  the   standard   quadratic  equation 
ax^  +  bx+  c  =  0,  (1) 

in  which  a  is  assumed  to  represent  a  positive  number  different  from 
zero,  may  be  obtained  as  follows  : 

From  (1)  we  may  obtain  the  equivalent  equation 

-      bx      —  c 
a        a 
hoc  oX  u 

Using  —  as  a  *'  finder  term,"  it  appears  from =  -— »  (2) 

hx 
that  the  square  may  be  completed  with  reference  to  x^  -] by  adding  the 

square  of  -—  • 

T     1  ^      bx      /bY      -c       b^  ,„v 

Accordmgly,  x^  +  -  +  (^-- j   =  —  +  ^,'  (3) 

Hence,  ^x  +  —  j   =-^^'  (4) 

Extracting  the  sc^uare  roots  of  both  members, 

b  /b^  —  4ac  ,_v 

Instead  of  writing  the  two  separate  linear  equations  which,  taken  to- 
gether, form  a  system  of  which  (5)  is  a  convenient  abbreviation,  we  may 
proceed  as  follows : 

Transposing  ^—  to  the  right  member  and  simplifying  the  radical,  we 

obtain, 

b     .     I 


x  =  —  --±-—  a/6'^  —  4  ac. 
2a      2a  ^ 


rTM       c  -b  ±  y^b^-4ac  .„. 

Therefore,  x  = ^ (6) 

A  a 

The  separate  expressions  for  the  value  of  x  may  be  obtained  by  using 
either  the  +  or  the  —  sign  before  the  radical. 

Accordingly,  from  (6)  we  obtain  the  following  set  of  two  equations  in 

which  the  value  of  x  is  expressed  in  terms  of  the  known  numbers  a,  b, 

and  c: 

_  &  +  ^^2  _4ac           .            -b-  ^/b-^  -4ac 
X  — 5 .     and    x  = \ (7) 


488  FIRST  COURSE  IN  ALGEBRA 

In  obtaining  these  solutions  we  have  employed  only  those  methods  of 
derivation  which  lead  to  equivalent  equations.  Hence,  the  "expressed" 
values  found  are  the  solutions  of  the  original  equation,  and  there  are  no 
others. 

By  using  the  double  sign  ±  before  the  radical,  as  in  (6),  we  can  verify 
the  two  results  at  the  same  time. 

Substituting  the  expressed  values  (6)  for  x  in  equation  (1),  we  have, 

f_  b  ±  V6'^-4ac\g,   ,l-h±  a/6^-4 


/-h±\/h'^-4.acy        (-b±  Vb^-4ac\ 


+  c  =  0 


b^T^  h^b'^  -4ac  +  5»~4oc      -2b^±  2  6^/6^-400      4ac_ 
4a  "^  4a  "^  Ta  ~ 


6^  T  2  6  ^62  -4ac  +  62-4ac-2  62±2  6^6=^- 4  ac  +  4  ac  =  0 

0  =  0. 

24.  After  a  given  quadratic  equation  has  been  reduced  to  the 
standard  form,  ax^  -\-  bx  -\-  c  =  0,  the  solutions  may  be  obtained 
immediately  by  substituting  in  the  general  solution  of  the  standard 

equation,  x  = ,  the  values  corresponding  to  a,  6, 

and  c,  respectively,  in  the  given  equation  reduced  to  standard  fonn. 

Ex.1.    Solve  a:2  +  6x- 216  =  0. 

Referring  to  the  standard  quadratic  equation,  ax^  +  6x  +  c  =  0,  we  find 
that  1  corresponds  to  a,  6  to  6,  and  —  216  to  c. 

Hence,  substituting  these  values  for  a,  6,  and  c,  respectively,  in  the  general 


-         ,  —  6  ±  V^   —  4  ac 

formula  x  = ^ , 

2  a 


_  6  ±  /i/62  -  4(-  216) 

we  have  x  = =^-^ — - — ^^ 

2i 


_  _6_±jv/36_+864 

~    .  2 

_  _  6  J:  30 

~         2 
That  is,  X  =  12,  and  x  =  —  18. 

These  values  are  found  by  substitution  to  satisfy  the  original  equation. 
Hence,  they  are  its  roots. 

Ex.  2.    Solve  6  a:2  -  5  a;  -  375  =  0. 

Referring  to  the  standard  quadratic  equation,  ax^  +  6a;  +  c  =  0,  it  appears 
that  6  corresponds  to  a,  —  5  to  6,  and  —  375  to  c. 


QUADRATIC  EQUATIONS  489 

Hence,  substituting  these  values  in  the  general  formula, 


h  ±  -Y/P-4ac 


we  have 


(-5)±  V(-5)^-4((i)(-375) 
2-6 


_  +  5  db  V25  +  9W0 
~  12 

_  4-  5  ±  95 
12 
That  is,  X  =  8|-,     and     a:  =  —  7^  • 

Both  of  these  values  satisfy  the  original  equation,  and  hence  they  are  its 
roots. 

Ex.3.     Solve  a:2+ll  =  8a:. 

Observe  that,  when  reduced  to  the  standard  form  x"^  —  8x  -\-  11=:  0,  the 
numbers  1,  —8  and  11  correspond  to  a,  b  and  c  respectively  in  the 
standard  quadratic  equation     ax^  -^  bx  +  c  =  0. 

Substituting  these  values  in  the  general  formula. 


,,  .                                 _(_8)±  V(-8)'-4-  11 
we  obtain  x  =  — ^^ ^^ 


_  +  8  ±  ^(J4  -  44 
~  2 

"        2 
_S±  2^5 
"         2 
That  is,  x  =  4±  ^/5. 

(Compare  the  solutions  above  with  those  shown  in  Ex.  2,  §  20,  and  also 
see  Fig.  9.) 

Ex.  4.    Solve  a;2  +  4  a:  +  5  =  0.     (See  §  9  ;  also  Fig.  3.) 

Referring  to  the  standard  quadratic  equation,  we  find  that  a,  b,  and  c 
represent  the  values  1,  4,  and  5  respectively  in  the  given  equation. 

Hence,  substituting  these  values  for  a,  b,  and  c  in  the  general  solution, 


,  ,  .                                       -  4  ±  V16  -4-5 
we  obtain  x  = ^^-^ 

2 


•190  FIRST  COURSE  IN  ALGEBRA 


That  is,  x  =  -2±  V-  1- 

Let  the  student  check  these  solutions,  using  the  double  sign  before  V— !• 

Ex.  5.   Solve  ax^-{a-b)x-b  =  0. 

It  will  be  observed  that  the  coefficient  a  of  x^  in  the  given  equation 
corresponds  to  the  coefficient  a  of  x^  in  the  standard  quadratic  equation 
ax^  -r  bx  -\-  c  =  0  ;  that  —  (a  —  b)  which  is  the  coefficient  of  x,  corresponds 
to  b  in  the  standard  equation  ;  and  that  the  known  term  —  6  in  the  given 
equation  corresponds  to  the  known  term  +  c  in  the  standard  equation. 

Hence,  substituting  these  values  in  the  general  formula, 


-b  ±  Wb*  —  4 ac 
x=i ^ 


we  have. 

[_  (a  _  6)]  ±  V[- (a  -  6)?  -  4a(- 6) 

2a 

+  (a  -  6)  ±  -y/itP'  ~2ab  +  b^  +  4ab 

~                              2a 

+  («  -  6)  ±  (rt  +  &) 
~                 2a 

Hence, 

-l-a  —  6  +  a  +  6                ,                -]-a  — b  — a  — b 

2a                     ^^^        ^=             2a 
2a                                                        -26 
2a                                                          2a 

.  =  -". 

a 

(Compare  this  method  of  solution  with  that  in  Ex.  29,  Exercise  XXII,  4.) 

Exercise  XXII.     6 

Solve  the  foUowing  equations  by  the  formula,  verifying  such  re- 
sults as  are  neither  irrational  nor  imaginary.  Whenever  surds 
appear  in  the  exact  solutions,  approximate  solutions  correct  to  four 
places  of  decimals  should  be  obtained. 

1.  a'  —  6  £c  +  5  =  0.  1,  x^—12x+  16  =  0. 

2.  ar^  +  9  «  =  —  20.  S.  x^—Ux+  d  =  0. 

3.  a;2  -  12  =  x.  9.  x" -\-  16  =  6x. 

4.  ic2  _  45  =  4a;.  10.  tc"  -  6 ic  +  1  ==  0. 

5.  Q^+  llx=12.  11.  £c'+  16  =  4a;. 

6.  a;^  +  8a;=l.  12.  4a;"' —  3a;  =  85, 


23. 


QUADRATIC  EQUATIONS  491 

13.  2aj'=5a;+  117.  ^,    x^       1       x 

24. —  =  -  • 

14.  a;-+  25  =  ISx.  21       7       3 

15.  6a;2+  7ic+  8  =  0.  25    -  -  -  +  i  =  0 

16.  7a;2-6£c  +  5  =  0.  '    '^       ^      '^ 

17.  x(x  +  12)  =  -  27.  26.  x"  -  2mx  +  n"  =  0. 

18.  15^(a.-l)  =  2(a.-2).  21.  x^  + Aab  =  2ax  +  2bx. 
/         X         /             .  28.  «(«;'  -  1)  =  (a'  -  l)x. 

19.  3a;(a;— 1)  =  2(1  +  2ic).  ^^  2   ,   /j      \7     i    i 
^          /         V     '        /  29.  acx^  -\-  bd  =  adx  +  hex. 

20.  (3a; -2)2  =  7a;.  ^2  ^  ^2 

21.  {x  -  2)2  +  5(a;  -  3)^  =  0.       ^0.  x" ^^  a;  +  1  =  0. 

22.  2a;2~  1.1  a;  =  4.2.  _ 

+  12^    ^_x  a  +  b       ^ 

12      ~^       2'  Z2.  x'  +  d^  +  ^cix-cTj^^^dx. 

33.  a^»ca;2  -  (a^^^  _|_  ^2^^^  _|_  ^^^  ^  q^ 

Rational  Fractional  liquations.  Containing  One 
Unknown 

25.  We  shall  now  consider  equations,  which  are  rational  and 
fractional  with  reference  to  a  specified  unknown,  the  solutions  of 
which  may  be  made  to  depend  upon  the  solutions  of  quadratic 
equations. 

26.  If  a  fractional  equation  cannot  be  solved  by  inspection,  then 
its  solution  may  be  made  to  depend  upon  the  solution  of  a  rational 
integral  equation.  This  integral  equation  may  be  derived  from  the 
given  fractional  equation  by  multiplying  the  terms  of  both  mem- 
bers by  the  lowest  common  multiple  of  the  denominators  of  all  of 
the  fractions  which  appear  in  the  different  terms  of  the  equation. 

27.  The  derived  integral  equation  will  not  always  be  equivalent 
to  the  given  fractional  equation,  for  in  exceptional  cases  it  may 
happen  that  extra  roots  which  do  not  satisfy  the  given  fractional 
equation  are  introduced  into  the  derived  equation. 

28.  Such  extra  roots  as  may  be  introduced  during  the  process 
of  clearing  a  given  fractional  equation  of  fractions  may  be  deter- 
mined by  an  examination  of  the  equation,  as  may  be  seen  by  con- 
sidering the  following 

Principle  Relating*  to  Extra  Roots  :  If  an  integral  equation 


492  FIRST  COURSE  IN  ALGEBRA 

is  derived  from  an  equation  which  is  rational  and  fractional  with 
reference  to  a  specified  unknoion^  x,  by  multiplying  both  members  of 
the  fractional  equation  by  a  function  of  x^  it  may  have  soluticms  which 
do  not  satisfy  the  given  fractional  equation. 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Let  all  the  terms  of  a  given  equation  which  is  rational  and  fractional 
with  reference  to  a  specified  unknown,  x,  be  transposed  to  the  first  member 
and  then  added  algebraically. 

Let  -y-  represent  the  resulting  rational  fractional  expression,  N  and  I) 

being  rational  integral  polynomials  with  reference  to  the  unknown,  x.    Then 

since  the  second  member  of  the  derived  fractional  ei^uation  is  zero,  it  follows 

that  the  given  fractional  equatioir  will  be  equivalent  to  the  derived  frac- 

N 
tional  equation  —  =  0  (1). 

N 
We  have  seen  in  Chapter  XVI,  §  4,  that  if  y-  be  reduced  to  lowest 

N 
terms,  the  fractional  equation     -  =  0  (1)  is   equivalent  to  the  integral 

equation  N  =  0  (2).     The  latter  is  derived  by  multiplying  both  members 

of    tj  =  0  by  the  multiplier  D,  which  is  necessary  in  order  to  clear  (1)  of 

fractions. 

N 
If,  however,  -=r  is  not  in  lowest  terms  with  reference  to  x,  then  the 

values  of  a;  which  reduce  the  factor  common  to  N  and  D  to  zero  will  reduce 

N 
both  the  numerator  and  denominator  of  the  fraction  -j-   to  zero,  and  for 

N 
such  values  of  x  the  true  value  of  jz  may  be  different  from  zero. 

Accordingly,  such   values  of  x  may  not  satisfy  the  fractional  equation 

Hence,  such  solutions  of  the  derived  integral  equation  N  =  0  as  are  solu- 
tions also  of  the  equation  constructed  by  jyla^ing  the  multiplier  D  equal  to  zero, 

may  not  be  solutions  also  of  the  fractional  equation  —  =  0,   and  accordingly 

must  he  rejected. 

Hence,  if  when  deriving  the  integral  equation  iV^  =  0,  a  multiplier  D  is 
used  which  contains  factors  which  are  not  necessary  to  clear  the  fractional 

N 
equation  —  =  0  of  fractions,  extra  solutions  may  be  introduced  by  the 

process. 


QUADRATIC  EQUATIONS  493 

12 

Ex.  1.   Solve + =  1. 

re  —  5       X  —  S 

The  fractions  are  in  lowest  terms  and  an  integral  equation  may  be  derived 

by  multiplying  the  separate  terms  by  (x  —  b){x  —  3)  wliich  is  the  lowest 

common  multiple  of  the  denominators  of  the  fractions  in  the  given  equation. 

Accordingly,         x  —  3  -\-  2(x  —  b)  =  (a:  —  5)  (a;  —  3). 

Reducing  to  standard  form,  we  obtain 

a;2_iia;4-28  =  0, 

the  solutions  of  which  are  found  to  be 

X  =  4,  and  x  =  7. 

It  should  be  observed  that  neither  of  these  values  is  a  solution  of  the 
equation  (x  —  5)(x  —  3)  =  0  which  is  obtained  by  equating  the  multiplier, 
(aj  —  5)(a;  —  3),  to  zero.  Hence,  neither  of  these  values  can  have  been  in- 
troduced as  an  extra  root  during  the  process  of  solution. 

The  values  4  and  7  are  found  by  substitution  to  be  the  solutions  of  the 
original  equation. 


Substituting  4, 

Substituting  7, 

z^-f-- 

M-- 

1  =  1. 

1  =  1. 

"Ry     9      Sr.lvp                            - ^ 

- 1 

-3  = 

1 

x^  - 

-1  " 

"a;-l 

(1) 

Observe  that  the  fractions  are  in  lowest  terms.  Using  as  a  multiplier 
cc^  —  1,  which  is  the  lowest  common  multiple  of  the  denominators,  we  may 
derive  the  integral  equation, 

3  a;  -  1  -  3(a;2  -  1)  =  a;  +  1.  (2) 

From  this  we  obtain  the  equivalent  equation  in  standard  form, 

3a;2_2a;- 1  =  0.  (3) 

Factoring,  (3  a;  +  l)(a:  -  1)  =  0.  (4) 

This  last  equation  is  equivalent  to  the  set  of  two  linear  equations  formed 
by  writing  the  factors  of  the  first  member  separately  equal  to  zero. 

3  a:  +  1  =  0,  and  a;  -  1  =  0.  (5) 

The  solutions  of  these  equations  are 

a;  =  —  ^,         and  a:  =  1. 

It  should  be  observed  that  the  value  a:  =  —  ^  is  not  a  solution  of  the 
equation  obtained  by  e(iuatii]g  the  multiplier  x^  —  1  to  zero.  Hence  it 
cannot  have  been  introduced  during  the  process  of  solution. 


494 


FIRST  COURSE   IN   ALGEBRA 


(-K 


By  substitution  a:  =  —  ^  is  fouud  to  be  a  solution  of  the  given  equation. 

(See  Fig.  12.) 

Y          :  The  remaining  value,    a:=l, 

/  is  a  solution  of  the  multiplier 

;  equation,  a:^  —  1  =  0,  and  accord- 

/  ingly  may  have  been  introduced 

•  as  an  extra  root  during  the  process 

^1  r\\  i?_^  -n      o^  solution. 

,£iliO]_Extra Root       ^  ,    ...     •         .     .      ,        , 

.•         jf  By    substitution   it   is   found 

that  this  value  does  not  satisfy 
the  given  equation,  and  accord- 
ingly it  must  be  rejected  as  being 
an  extra  root.     (See  Fig.  12.) 

Hence,  x  =  —  \  is  the  single 
solution  of  the  given  equation. 

If,  instead  of  deriving  an  inte- 
gral equation  from  the  fractional 
equation  in  the  form  as  given, 
we  first  write  equation  (1)  in  the 
form 


(  Dotted  line  3x2 -2t 
3a:-  1 


3  =  .y. 
z  +  1 


a:«-  1 


Combining  the  fractions, 


which  reduces  to 


{X  -  l)(ar  +  1) 

2(^  -  1) 
(X  +  l)(a:  -  1) 
2 


a;2  _  1      a;  -  1 
we  may  obtain 
3  =  0. 


0, 


3 


a:+l 
Hence,  2  -  3  x  -  3  =  0. 

The  single  solution  of  this  last  equation  is  a:  =  —  ^. 

It  will  be  observed  that,  when  deriving  the  integral  equation  in  the  first 
solution,  we  used  a  multiplier  containing  more  factors  than  were  necessary 
to  clear  of  fractions,  and  accordingly  introduced  an  extra  solution  into  the 
derived  equation. 

29.   It  is  often  advisable,  before  deriving  an  integral  equation,  to 
combine  the  fractions  in  a  given  fractional  equation,  and  to  reduce 
to  lowest  terms  all  of  the  fractions  then  appearing. 
1111 


Ex.3.   Solve 


+ 


+ 


a;-3a;-7      a;  +  2a;-2 


QUADRATIC  EQUATIONS  495 

Instead  of  deriving  an  integral  equation  at  once  by  multiplying  the  terms 
by  the  lowest  common  multiple  of  the  denominators,  we  shall  find  it  to  our 
advantage  to  transpose  the  fractional  terms  in  such  a  way  as  to  obtain  differ- 
ences between  two  pairs  of  fractions,  as  follows  : 


X  —  3      x  +  2      a;— / 
Combining  the  first  and  second  fractions,  and  also  the  third  and  fourth 
fractions,  we  obtain 


(x-3)(x-\-2)   '   (a:-7)(x-2) 
Multiplying  the  terms  of  this  equation  by  the  lowest  common  multiple 
of  the  denominators,  (x  — '3)(x  -\-  2)(x  —  l)(x  —  2),  and  dividing  both  mem- 
bers by  5,  we  obtain 

{x  -  7)(x  -  2)  +  (x  -  3)(x  +  2)  =  0, 
which  reduces  to  a:^  —  5  a;  +  4  =:  0, 

The  solutions  of  this  equation  are 

a:  =  4,  and  a;  ~  1. 
Neither  of  these  values  is  a  solution  of  the  equation 
(x  -  3)(x  +  2)(x  -  l){x  -  2)  =  0 
obtained  by  equating  to  zero  the  multiplier  used  in  deriving  the  integral 
equation  a;^  —  5  a;  +  4  =  0.     Hence  neither  value  can  be  an  extra  root  intro- 
duced during  the  process  of  solution,  and  accordingly  these  values  must  be 
solutions  of  the  given  equation. 

Both  values  are  found  by  substitution  to  satisfy  the  given  equation. 

^     ^    g^^^^     ^ x-Q  _  _x^ 2  (a;  -  6)       ^ 

a:  —  2       (a;  —  4)(a;  —  2)  "~  a;  —  3       (a;  —  5)(a:  —  3) 

Clearing  of  fractions  by  multiplying  all  of  the  terms  by  the  lowest  com- 
mon multiple  of  the  denominators,  (a:  —  2)(x  —  4)(a5  —  3)  (a;  —  5),  and  com- 
bining terms,  we  obtain  the  integral  equation  a;^  _  5  ^  —  6  =  0,  the  solutions 
of  which  are  found  to  be  a;  =  3,  and  a:  =  2. 

Since  these  values  are  solutions  of  the  equation 

{x  -  2)(x  -  4)(a;  -  3)(a;  -  5)  =  0, 

formed  by  placing  equal  to  zero  the  multiplier  employed  in  clearing  the 
given  equation  of  fractions,  they  must  be  rejected  as  being  extra  roots 
introduced  during  the  process  of  solution. 

By  substituting  these  values,  it  will  be  found  that  neither  satisfies  the 
original  equation. 

Accordingly  the  fractional  equation,  as  given,  has  no  finite  solution. 


496                        FIRST  COURSE  IN  ALGEBRA 

Exercise  XXIL     7 

Solve  the  following  equations,  rejecting  extra  solutions,  and  veri- 
fying all  others  : 

^      2^a      8  _4 24       ^ 

2"'"a;      8a;  'ic-6      a;  (a; -6) 

7           6    _13  7a;+23_8a;  +  7 

^-  6^""7a;^~42'  ^^-  ^        a;+ 2   ~  a;+ 2  ~^^- 

q    l4.A=l?  17    g  +  5      a;  -  5  ^  37 

ar»'^5«^       5  a;-5"^a;+5        6 

a;-6^      8  a;  +  4      a;-4^82 

8      ~a;+6*  •a5-4a;  +  4~9' 

^*       4      -a:-3  ^^^    x' - 'd       a;  -  3       ^  "  ^• 

6.  7  =  a.  20. —  1  = 


a;  —  4  aj— 1  a;^— 1 

2a;  +  1  ^       1  G  +  a;      a;  -  5  ^  15 

3  2a;- 1*  *  l+a;         2a;         8  * 

a;  —  5       a;  a;--2a;(a;  —  2) 

1— a;  1— a;  a;  —  2      a;^  —  4 

=  9.  24.^+      2  1 


10. 

X 

+ 

3a;- 
a;  - 

-8 
-3 

X  - 

-7 
-3 

5a;- 

-7 

2a;- 

■21 

XX. 

ji. 

X  - 

-4 

a;  •— 

4 

12. 

1 

X 

+ 

1 
13' 

1 
a;-f  13 

13. 

X 

X 

T^ 

i- 

a;-f  1 

X 

• 

14. 

X 

7a;- 

-30 

_3a;- 

20 

a;— 1       a;+2       a;-Fl 

1.  25.^  +  ^  =  6. 

a;  —  2         a;^  —  4 

26.^  +  1  =  -?^. 

X  +  2       a;      a;  -f  3 

27. 1 + -4— + '-4— = ^• 

3       S  +  X      4  +  a; 

2.  28.^+      ^  « 


a;— 5  a;  — 5  a;  —  1      a;  —  2      a;  —  3 


QUADRATIC  EQUATIONS  497 

3 


zy. 

a;  — 4       (x—l)(x'-4:)       a;  +  6 

30. 

1                      11                         1 

a;  —  5       (x+  Q)(x  —  6)  ~  S  (x  —  6) 

31. 

2a;-  7               2a;+  3       17 
a;-l     '       ~       a;            20 

32. 

4(a;+  5)_7a;-  10 
a;  —  3                7 

33. 

2a;+3       7a;— 6 

a;— 5         a;+  7 

34. 

'      +      ^'     =1. 

a;  +  3       a;^  -  9 

35. 

5               2           1 

^-l       a;  +  1  ~  8 

36. 

7           17a;- 155  _„ 
a;  — 5          x^  —  2b 

37. 

X           a;  —  3       5 
a;  -  1    '   a;^  -  1  ~  4 

38. 

5          4       17  a;  — 42 
a; -6       5~    a;^  -  36 

39. 

1                14            1 

a;  -  7       a;2  -  49  ~  17 

d.n 

X          9(a;  -  1)       2 

a;-3         a;2-9~5 

41. 

1                 1                  1 

1— a;       1  —  2a;       1  —  3  a; 

42. 

10           9              8 
X       a;  +  1       a;  +  2 

43. 

4              3                1 

4-a;       3  — a;       1  -4aj 

44. 

5               3              16 

5  —  a;      3  —  a;      a;  —  2 

45. 

a;  —  1       5            5  —  x 

x—S       2  ~  a;2  -  5 a;  -f-  6 
32 

498  FIRST  COURSE  IN  ALGEBRA 

_     X—5X  +  4:         25 
4b. + 


47. 

48. 


X—  7       2aj+  5        7 
1  aj+ll 


25-2       (2aj+ 9)(a;-2)       3(2a;-5) 
1  x+1  3 


3iC-2       (4aj- l)(3ic-2)       5(2 a;  +  1) 


49  ^  325+2^9 

'   2(a;  +2)       aj*  -  4       10 

50.^+  ^-^  ^ 


$1. 
52. 
53. 
54. 
55. 


aj— 2  (x—l)(x—2)      x+1 

x  —  4:  3a;— 12_      5 

x—d  x  +  4:    ~~  X—  d' 
2  a;- 3  5 


a;  +  5  (a;  +  l)(x ,+  5)  ~  a;  +  9 
1 7  2 

a;  +  -4  (a;  -  3)(a;  +  4)  ~  a;  +  2 
2  x+  S  1 


a; +5       (a;+4)(a;+5)       4(a;  -  8) 

2 a;  — 8       _      2 

a;+  2      ar'  — a;--6""a;+  l' 


56.   -L+  '  ' 


57. 

58. 


a;  — 5      ar^— 11  a; +30      a;  +  10 
1  1  1 


a; +6       2(aj-3)      a^+lSx  +  4.2 
1  3  1 


2a;-l       (4a;+  l)(2a;- 1)       8a;-l 


59.   -L^H-       '  ' 


a; -4       3a;- 7       (2a;  -  3)(a;  —  4) 

5 26 1 

5  a;  —  1       (x+  5)(5  a;  -  l)  ~"  3  a;  —  5 

61.    ,r^+  ''  ' 


60. 


6  a;— 1       (5  a;  —  9)(6  a;  —  1)       a;  +  3 
62.-^+         2a;- 10  4 


3a;— 1       (a;+  l)(3a;— 1)       3a;+l 


63. 


QUADRATIC  EQUATIONS  499 

1  X-  16  2 


3ic— 2       (4a3  +  3)(3a3  — 2)       6  a;— 1 


64.        ^''       +      4(/>..-ll)       _       15 


5  a;  —  2       (a;  +  2)(5  a;  —  2)       3  a;  +  2 
65.         ^  2(5  a;  4- 2)  _       1 


66. 
67. 

68. 


2  a;  — 5       (12  a;  —  l)(2a;  —  5)       4  a; +5 

1 4(a;+  1)  __         3 

2a;  +  3       H^x-\-  7)(2a;+  3)  ~  6a;+  u' 
X  5x  —  6  x^-7  3 


a;  — 6       (a;  — 2)(a;  — 6)       (a;  -  3)(a;  —  1)       a;  —  1 

a;- 5  _         a;+  17         _  a;^  — 5a;—  15 1_ 

a; -7       (x  +  5)(x  -  1)  ~  (x  -  8)(a;  +  1)       a;  +  l" 


69.   -^+, ^ .=^-+         '"-' 


70, 


aj-1    '    (^  — 3)(a;— 1)       a;  +  1       (a;  — 2)(a;+l) 
X  13  a; +24  x  7  a; +8 


a; -4       (a;+ll)(a;  — 4)       a;  —  2       (a;  +  9)(a;  -  2) 

X 2  (a;  —  9)       _      x a;  —  8 

*   a;  —  3       (x—  l){x  -  3)  ~  a;  —  2       {x  —  b)(x  —  2)  ' 

X 3  (5 a;  +  8)       _  _x 5a;  +  4 

■    a;  —  6       (a;  +  13)(a;  —  6)  ~  a;  +  4       (x  +  8)(a;  +  4) 
-^         ^       .        12  (a;  +  2)      _      x  Ix  —  20 


a; +3       (a;— l)(a;+3)       a;  —  2       (a;  +  iX^c  —  2) 
74.        ^       I         4  (a; -5)       ^      x     ^         6  (a;  -  3) 


a;  — 4       (a;  — 3)(a;  — 4)       a;  —  2       (a;  +  l)(a;  —  2) 

X  2(3a;+8)      _      x  7a; +15 

a;  +  2  "^  (a;  +  4)(a;  +  2)  ~  a;  +  3  "^"  (a;  +  l)(a;  +  c 


3) 


Exercise  XXII.   8.     Miscellaneous 


Solve  the  following  equations,  verifying  such  solutions  as  are 
neither  irrational  nor  imaginary  : 

1.  a;'  —  15a;  +  54  =  0.  4.   (x  +  4)^  =  9x^ 

2.  2  a;^  —  7  a;  =  15.  5.  55(a;''^  —  a;)  =  11  a;. 

3.  3(a;2  -  1)  =  8  x.  6.   (x  +  2)(a;  +  3)  =  20. 


500  FIRST  COURSE  IN  ALGEBRA 


7.  (a;+ l)(a;-4)  =  50.  22.  i  -  ^  =      ^ 


28. 

2x 
3 

..•=!■ 

29. 

15  a; 

X-  1 

=  ll(a;- 

1). 

Qfi 

1 

1 

1 

X       2       x  —  2 

s-^ZTT-iO'^-a-  23.  -  +  -  =  -. 

2ar  —  1  7        a;      5 

,0.^+1  =  2  +  1.  25.?-^  =  ^-5, 

2  a;  5       6       5a; 

n.. +  1=5  +  1.  26.^  +  ?  =  ^. 

5  a;  2        a;        3 

12.   10(aH-2)(a^-2)  =  41a;.  27.  -  +  a;  =  3  +  -• 

X  X 

IS.  x^  +  x  =  (P  +  d, 
U.  x-{-\  =  b-h-' 

0  X 

mn 

15.  X  -\ =  m  -\-  n. 

X  a;+  1      a;+2      a;+3 

-./>      2  .     2         2  .   ^  o^    « +  16       64  — a;      x 

16.  x^  +  n^  =  m^  +  2  nx.  31.  — =  -  —  13. 

8  a;  —  4       4 

17.  a;(2a;-«)  +  a;(a;-a)=^>a;.     32.  -^  +  ^^  =  i^. 

a;  +  1  a;  6 

^^-  2  ^  a;  ~  2  ^  3  '^'^'  x      6  ~  a;  -  6 

19.?4-^  =  -^  +  ^  34.  U-^  =  i. 

a      X      a      0  a;8— a;8 

20  _^  +  _?--3  35    -^-  +  ^±1-11^. 
^"-  a;  -  3  +  a;  +  2  ~  •^"  ''•^-  a;  +  1  +      x      ~  56 

21  _3 5^_^^.  36    ^±2   ,   ^±3^41. 

a;— 5  a;-3  a;  +  3      a:  +  2      20 

a;  —  2  a;^  —  4       a;+2 

38.  (2a;  +  l)(2a;  -  1)  +  (a;  +  2)(a;  —  2)  =  x{x  +  3). 


QUADRATIC  EQUATIONS  501 

39.    (x  +  2)(x  -  3)  +  (^  +  9)(a;  +  3)  =  3aj  +  15. 

X      a      X  —  a 

41.  «      +_:^  =  2. 
X  —  0      X  —  a 

•  ^    a   ,   £c      33a^  — £c^ 

42.  — I —  = 

X      a  ax 

a;       a^      aj       6^ 

43.  -2  H —  =72  H 

a^      X       Ir      X  k 

44.  a;2+  8a;=  16^  +  2^a;. 

45.  a'  +  2^;a;  =  rt2_|.  2^^,. 

46.  {x  +  />)2  +  6(a;  +  ^)  +  9  =  0. 

47.  a;'  +  4«6  =  2ic(^)  +  a). 

48.  2a;(7ic- «)  =  («  +  «)(« -a). 
7W  +  1 


49.   ^  = 
50. 


x^       a;  +  1 

a  —  4w  +  w       9  n  +  x 


n  —  m  X 

^^     (x+l)(x-2)       (x-l)(x-h2)  ^ 
2  3 

52.  (a  -  xy  +  (b-  xY  =  (b-  ay. 

53.  (x  -  gY  +  {x-  hY  =  /  +  h''. 

54.  (c  —  a)x^  +  {a  —  b)x  +  b  —  c  =  0. 

a  +  b  +  X      a      b      x 

56    ^  +  ^      ^  ~  ^  ^  1       ^'  +  2  <^ 
'   X  —  d      X  +  d  x^  —  d^ 


a;      a;  +  4       aj+1       7a;  +  3 

',8.        ^       I       ^     ^      ^       I       ^ 
*6  +  iC      c  +  a;      a  +  6      a  +  c 


502  FIRST   COURSE   IN   ALGEBRA 

Problems  Solved  by  Means  of  Quadratic  Equations 

30.  In  the  solution  of  problems  by  means  of  conditional  equations, 
it  is  often  convenient  to  represent  the  unknown  quantities  by  such 
"initial  letters  "  as  may  suggest  the  quantities  considered. 

E.fj.  In  the  case  of  a  body  movinjjj  at  a  uniform  rate  during  a  specified 
time,  the  distance  passed  over  may  be  found  as  follows : 

distance  =  rate  x  time. 

In  particular,  if  a  train  moves  uniformly  at  the  rate  of  forty  miles  an 
hour  for  two  hours,  the  distence  travelled  will  be  eighty  miles. 

If  we  represent  the  number  of  units  in  terms  of  which  distance,  rate,  and 
time  ai-e  expressed  by  the  initial  letters  d,  r,  and  t,  respectively,  we  may 
express  the  general  relation  between  distance,  rate,  and  time  for  uniform 
motion  by  the  formula  d  =  r  x  t. 

For  simple  interest  we  have  the  relation 

interest  =  principal  x  Tate  x  time. 

Representing  the  numbers,  in  terms  of  which  interest,  principal,  "rate,  and 
time  ai-e  expressed,  by  the  initial  letters  i,  p,  r,  and  t,  respectively,  we  have 
the  formula  i  =  p  x  r  x  t. 

31.  When  using  a  particular  formula,  the  letters  which  are  to  be 
regarded  as  representing  unknowns  must  be  determined  by  the 
nature  of  the  problem  to  the  solution  of  which  it  is  applied. 

Solution  of  Formulas  for  Specified  Letters 

32.  The  Greek  letter  tt,  read  "pi,"  is  used  in  mathematical 
calculations  to  represent  a  certain  incommensurable  constant  the 
approximate  value  of  which  may  for  many  practical  purposes  be 
taken  equal  to  22/7.  In  particular,  the  ratio  of  the  circumference 
to  the  diameter  of  a  circle  is  equal  to  tt. 

Exercise  XXII.     9 

In  the  following  formulas  find  the  expressed  values  of  the  letters 
specified  in  terms  of  those  remaining: 

1.  Solve  for  E,     s  =  ttR''.  3.  Solve  for  t,       S=  ^at\ 


m 


ms 


2.  Solve  for  E,     w  =  —^  •  4.  Solve  for  s,      F  = 


PROBLEMS  503 


5.  Solve  for  r,       /=  — ^• 

r 

a^    - 

6.  Solve  for  a,       s  =  —  ^3. 

4 

7.  Solve  for  n,      d  =  -—- ^  • 

n(n  -  1) 

Find  the  value  of  n,  \{  d  =  2,  /  =  19,  and  s  =  96. 

8.  Solve  for  ^,     V  =  ^  ttE^. 

Find  the  value  of  ^,  if  T  =  132,  and  h  =  14. 

9.  Solve  for  R,        T  =  2  tt  R  (ff  +  R). 

Find  the  value  of  i^,  if  r=  352,  and  j^  =  10. 

10.  Solve  for  t,  ab  =  t^  +  pq. 

Find  the  value  of  #,  if  a  =  16,  6  =  14,  jo  =  7,  and  q  = 

11.  Solve  for  c,  a^  =  b^  +  c^  +  2  cp. 

Find  the  value  of  c,  if  «  =  18,  6=11,  and  jt?  =  |^. 

12.  Solve  for  m,       2  a""  +  2  ^^^  =  c^  +  4  m^ 

Find  the  value  of  m,  if  a  =  12,  b  =  16,  and  c  =  20. 

13.  Solve  for  n,        s  =  ^[2a  +(?^  —  l)d]. 

Find  the  value  of  n^  if  5  =  272,  a  =  6,  and  c?  =  8. 

14.  Solve  for  «,         s  =  — h  • 

2  2t  db 

Find  the  value  of  a,  if  5  =  70,  /  =  16,  and  d  —  2. 

V"  —  a^ 

15.  Solve  for  /,  d  = -, 

2s— I  —  a 

Find  the  value  of  /,  if  c?  =  15,  a  =  18,  and  5  =  1206. 


The  Solution  of  Problems 

33.  Whenever  the  translation  into  algebraic  language  of  stated 
relations  between  the  unknown  quantities  of  a  problem  leads  to  one 
or  more  conditional  equations  of  the  second  degree,  we  may  expect 
to  find  two  or  more  solutions. 

It  may  happen,  however,  that  one  or  both  of  the  solutions  of  the 
algebraic  equations  must  be  rejected  as  not  fulfilling  the  conditions 
expressed  by  the  given  problem. 


504  FIRST  COURSE   IN  ALGEBRA 

Solutions  consisting  of  negative  numbers  or  fractions  cannot,  in 
certain  cases,  be  given  a  sensible  interpretation. 

E.  g.  If  the  number  of  people  in  a  given  assembly  be  in  question,  negative 
or  fractional  answers  must  necessarily  be  rejected. 

In  the  case  of  a  body  in  motion,  fractions  have  a  significance  aa  indicat- 
ing definite  distances,  while  negative  answers  may  be  interpreted  as  meaning 
a  reversal  in  the  direction  of  the  motion. 

34.  Positive  results  will  in  general  be  found  to  satisfy  all  the 
conditions  of  a  given  problem. 

A  negative  result  will,  in  general,  satisfy  the  conditions  of  such 
problems  as  are  concerned  with  abstract  numbers. 

Whenever  the  unknown  quantities  referred  to  in  a  problem  are  of 
such  nature  as  to  admit  of  being  taken  in  opposite  senses,  that  is,  of 
being  regarded  as  positive  or  negative,  then  in  such  cases  it  is 
usually  possible  to  give  a  sensible  interpretation  to  a  negative  result. 

Ex.  1.  Find  two  numbers  of  which  the  sum  is  30,  and  of  which  the 
product  is  221. 

If  X  stands  for  one  of  the  required  numbers,  then  by  one  of  the  condi- 
tions of  the  problem,  30  —  a:  will  stand  for  the  other. 

We  may  express  the  remaining  condition,  that  the  product  is  221,  by 
means  of  the  conditional  equation 

a:(30-a:)  =  221. 

Solving,  a:  =  17,         and        a;  =  13. 

Hence  one  of  the  numbers  is  17  or  13,  and  substituting  either  of  these 
values  for  x  in  the  expression  for  the  other  number,  30  —  a:,  we  obtain  13 
and  17  respectively. 

These  values  satisfy  both  the  algebraic  equation  and  also  the  conditions 
of  the  problem  as  stated,  and  hence  are  its  solutions. 

35,  Imaginary  results  must  always  be  interpreted  as  indicating 
inconsistent  relations  among  the  conditions  of  a  given  problem  as 
stated. 

Ex.  2.    Separate  50  into  two  parts  the  product  of  which  is  630. 

We  may  represent  the  required  numbers  by  x  and  50  —  a:  respectively. 

Expressing  the  condition  that  the  product  of  these  numbers  is  630,  we 
may  construct  the  conditional  equation  x(50  —  x)  =  630,  the  solutions  of 
which  are  both  found  to  be  imaginary. 

These  imaginary  results  must  be  interpreted  as  indicating  that  two  such 
numbers  cannot  exist. 


PROBLEMS  505 

36.  When  expressing  the  relations  of  a  given  problem  by  means 
of  one  or  more  conditional  equations,  all  that  we  know  at  the  outset 
regarding  the  number  of  the  solutions  is  that  they  must  be  found 
among  the  abstract  numbers  which  constitute  the  algebraic  solu- 
tions of  the  equations  ;  if  none  of  the  numbers  thus  found  can  be 
given  a  reasonable  interpretation,  then  the  given  concrete  problem 
as  stated  has  no  solutions. 

Exercise  XXIL     10 

Solve  the  following  problems,  employing  equations  containing  one 
unknown  quantity  : 

1 .  Separate  42  into  two  parts  such  that  one  part  is  the  square  of  the 
other. 

2.  The  sum  of  two  numbers  is  45  and  their  product  is  476,     Find  the 
numbers. 

3.  Find  two  consecutive  integers  the  product  of  which  is  702. 

4.  Find  two  consecutive  even  integers  the  product  of  which  is  528. 

5.  Find  two  consecutive  odd  integers  the  product  of  which  is  1023. 

6.  Find  two  consecutive  integers,  the  sum  of  the  squares  of  which  is  481, 

7.  Find  two  consecutive  even  integers,  the  sum  of  the  squares  of  which 
is  2180. 

8.  Find  a  number  which  equals  4  more  than  the  square  of  its  fourth  part. 

9.  Find  three  consecutive  even  integers  such  that  the  product  of  the  first 
and  third  shall  equal  three  times  the  second, 

10.  Find  three  consecutive  even  integers  such  that  the  square  of  the 
greatest  shall  be  equal  to  the  sum  of  the  squares  of  the  other  two. 

11.  Find  three  consecutive  integers,  the  sum  of  the  products  of  which, 
by  pairs,  is  674. 

12.  Find  four  consecutive  even  integers  such  that  the  product  of  the  first 
and  third  shall  equal  the  sum  of  the  second  and  fourth. 

13.  Find  four  consecutive  odd  integers  such  that  twice  the  product  of 
the  second  and  fourth  is  equal  to  eleven  times  the  sum  of  the  first  and 
third. 

14.  Find  three  consecutive  integers  the  sum  of  which  is  one-third  the 
product  of  the  first  two. 

15.  Find  two  consecutive  integers  such  that  the  sum  of  their  reciprocals 
is  if. 

16.  Separate  a  number  represented  by  a  into  two  parts  such  that  one 
part  shall  be  the  square  of  the  other. 


506  FIRST  COURSE  IN  ALGEBRA 

17.  Find  a  number  such  that,  if  87  be  subtracted  from  it,  the  remainder 
equals  the  quotient  obtained  by  dividing  270  by  the  number. 

18.  Find  a  positive  fraction  such  that  two  times  its  square  is  3  more  than 
the  fraction. 

19.  The  denominator  of  a  certain  fraction  exceeds  its  numerator  by  3, 
and  the  reciprocal  of  the  fraction  exceeds  the  fraction  by  39/40.  Find  the 
fraction. 

20.  The  numerator  of  a  certain  improper  fraction  exceeds  the  denominator 
by  5,  ami  the  fraction  exceeds  its  reciprocal  by  45/14.     Find  the  fraction. 

21.  Find  two  numbers  in  the  ratio  2  :  7,  the  sum  of  the  squares  of  which 
is  212. 

22.  Find  a  number  such  that  the  sum  of  its  third  part  and  its  square  is 
1100. 

23.  Find  a  number  such  that  one-half  its  square  shall  exceed  the  square 
of  one-half  the  number  by  one-lialf  the  number. 

24.  If  the  seventh  part  and  the  eighth  part  of  a  certain  number  are  mul- 
tiplied together,  and  the  product  is  divided  by  3,  the  quotient  is  298|. 
Find  the  number. 

25.  It  is  found  that  when  a  number  which  is  the  product  of  three  con- 
secutive integral  numbers  is  divided  separately  by  each  of  these  three  factors, 
the  sum  of  the  quotients  thus  obtained  is  191,     What  are  the  numbers  ? 

26.  I  have  thought  of  a  number.  I  multiply  it  by  2^,  then  add  4  to 
the  product ;  I  then  multiply  the  result  by  three  times  the  number  thought 
of,  and  finally  divide  by  5  and  subtract  from  the  quotient  five  times  the 
number  originally  thought  of,  obtaining  thus  20.  What  was  the  original 
number  ? 

27.  It  is  necessary  to  construct  a  coal  bin  to  hold  6  tons  of  coal.  Allow- 
ing 40  cubic  feet  of  space  per  ton  of  coal,  what  must  be  the  dimensions  if, 
the  depth  being  6  feet,  the  length  is  equal  to  the  sum  of  the  width  and  the 
depth  ? 

28.  A  crew  can  row  8  miles,  down  a  stream  and  back  again,  in  3  hours 
and  40  minutes.  If  the  rate  of  the  stream  is  two  and  one-half  miles  an 
hour,  find  the  rate  of  the  crew  in  still  water  in  miles  per  hour. 

29.  A  man  bought  two  farms  for  $3600  each.  The  larger  contained  15 
acres  more  than  the  smaller,  but  $8  more  per  acre  was  paid  for  the  smaller 
than  for  the  larger.     How  many  acres  did  each  contain  1 

30.  If  $3000  amounts  to  $3213.675  when  put  at  compound  interest  for 
two  years,  interest  being  compounded  annually,  what  is  the  rate  per  cent 
per  year  ? 

31.  If  $4250  amounts  to  $4508.825  when  placed  at  compound  interest 
for  2  years,  interest  being  compounded  annually,  find  the  rate  per  cent 
per  year. 


PROBLEMS  507 

32.  Telegraph  poles  are  placed  at  equal  intervals  along  a  certain  railway. 
In  order  that  there  should  be  two  less  poles  per  mile  it  would  be  necessary 
to  increase  the  space  between  every  two  consecutive  poles  by  24  feet.  Find 
the  number  of  poles  to  a  mile. 

33.  A  man  bought  a  certain  number  of  shares  of  a  railway  stock  for 
16600.  The  next  day  they  declined  in  value  $12  a  share,  when  he  could 
have  bought  five  shares  more  for  the  same  amount.  Find  the  price  paid 
per  share. 

34.  A  broker  purchased  a  certain  number  of  shares  of  stock  for  $2560. 
After  reserving  10  shares,  those  remaining  were  sold  for  $2450  at  an  advance 
of  $3  a  share  on  the  cost  price.     How  many  shares  did  he  buy  ? 

35.  It  is  desired  to  carpet  a  floor  in  the  form  of  a  rectangle  15  feet  long 
by  12  feet  wide,  with  a  carpet  having  a  plain  color  border  of  uniform  width. 
Allowing  $1.44  per  square  yard  for  the  center  and  $0.45  per  square  yard 
for  the  border,  determine  the  width  of  the  border  in  order  that  the  entire 
expense  may  be  $18.68. 

36.  Two  steamers  ply  between  two  ports,  a  distance  of  475  miles.  One 
goes  half  a  mile  an  hour  faster  than  the  other,  and  requires  two  and  one- 
half  hours  less  for  the  voyage.  Find  the  rates  of  the  steamers  in  miles 
per  hour. 

Let  X  represent  the  rate  of  the  slower  steamer  in  miles  per  hour. 

Then  the  rate  of  the  faster  boat  in  miles  per  hour  will  be  represented 
by  X  +  1/2. 

From  the  general  formula  expressing  the  relation  between  distance,  rate, 
and  time  for  uniform  motion,  we  have,  by  the  conditions  of  the  problem, 
475/x  and  475/(a:-f  ^)  as  representing  the  times  required  for  the  slower 
and  faster  boats  respectively  to  make  the  entire  trip. 

By  the  conditions  of  the  problem,  the  time  required  by  the  faster  boat  is 
^\  hours  less  than  that  taken  by  the  slower.  Hence  we  obtain  the  condi- 
tional equation  475/x  =  475/(x  +  ^)  +  4. 

From  this  equation  we  obtain  the  integral  equation 
2ic2  +  ic- 190  =  0, 
the  solutions  of  which  are  found  to  be 

X  =  ^-,     and     x  =  -  10. 

Since,  from  the  nature  of  the  problem,  the  forward  motion  only  of  the 
boats  is  considered,  we  shall  use  the  positive  value,  x  =  9^,  and  reject  the 
negative  value  x  =  —  10, 

Accordingly,  x  +  1/2,  which  is  the  rate  of  the  faster  boat,  is  10  miles 
an  hour. 

The  values  9^  and  10  will  be  found  to  satisfy  the  condition  of  the  given 
problem. 


^08  FIRST  COURSE   IN  ALGEBRA 

37.  An  engineer,  on  a  trip  of  108  miles,  found  it  necessary  at  the 
thirty-sixth  milestone  to  decrease  his  speed  to  a  rate  9  miles  an  hour  less, 
with  the  result  that  he  was  on  the  road  24  minutes  longer  than  would  have 
been  the  case  had  no  alteration  in  speed  been  made.  Find  the  rate  in  miles 
per  hour  before  the  speed  was  changed  and  also  the  time  required  to  make 
the  entire  trip. 

Let  the  speetl,  in  miles  per  hour  before  the  change  was  made,  be  repre- 
sented by  X. 

By  the  conditions  of  the  problem  the  time,  36/a;,  required  to  cover  the 
first  36  miles,  taken  together  with  that  for  the  remaining  72  miles  at  the 
decreased  speed,  a;  —  9,  that  is,  72/(a;  —  9),  is  equal  to  the  time  which  would 
have  been  required  to  run  the  entire  distance  of  108  miles  at  x  miles  per 
hour,  —  that  is,  108/x,  increased  by  24/(jO  hours. 

Hence  we  have  the  conditional  equation 

36         72     _  108      24 
X  "^a:-9~T~'^60* 
which  reduces  to  x^  —  ^x  —  1620  =  0. 

Of  the  two  solutions  of  this  equation,  a:  =  45  and  x  =  —  36,  the  negative 
value  cannot  be  admitted,  since  only  the  forward  motion  of  the  train  is  in 
question. 

Hence,  as  a  solution  of  the  problem,  we  find  that  at  first  the  train  was 
going  at  the  rate  of  45  miles  per  hour. 

The  time  in  hours  required  to  cover  the  entire  distance  is  expressed  by 
36/a; +  72/(3; -9). 

Substituting  45  for  x  in  this  expression  we  obtain  as  the  number  of  hours 
required  for  the  entire  trip,   ff  +  H»   which  reduces  to  2-|. 

These  values,  45  and  2|,  will  be  found  to  satisfy  the  conditions  of  the 
given  problem. 

38.  In  answering  a  false  alarm,  a  fire  engine  travelled  a  distance  of  3/5 
of  a  mile  at  the  rate  of  5  miles  an  hour  faster  than  when  returning.  If  it 
returned  immediately  on  reaching  the  "alarm  box"  and  was  gone  from  the 
station  16^  minutes  in  all,  what  was  its  rate  at  first  in  miles  per  hour  ? 

39.  A  and  B  run  a  half-mile  race.  A,  who  is  faster  than  B  by  1/2  a 
yard  a  second,  allows  B  a  start  of  1/4  of  a  minute,  and  beats  him  by  5 
yards.     Find  their  respective  rates  in  yards  per  second. 

40.  A  warship  which  is  approaching  a  port  is  discovered  when  it  is  12 
miles  away.  A  flotilla  of  torpedo  boats,  the  maximum  speed  of  which  is 
known  to  exceed  that  of  the  warship  by  7  miles  an  hour,  is  sent  out  5f 
minutes  later  to  meet  the  warship.  They  intercept  it  when  it  has  covered 
half  the  distance  to  the  port.     Find  the  rate  of  the  warship  in  miles  per  hour. 


PROBLEMS   IN   PHYSICS  509 

41.  After  having  gone  40  miles  of  his  trip  at  a  uniform  rate,  an  engineer 
found  that  his  train  was  behind  time.  He  immediately  increased  the  speed 
of  the  engine  to  a  rate  4  miles  an  hour  more,  and  completed  the  trip  of  62 
miles,  arriving  at  the  terminus  3  minutes  earlier  than  would  have  been  the 
case  if  no  change  in  rate  had  been  made.  Find  the  rate  of  the  train  in 
miles  per  hour. 

42.  A  man  travels  24  miles  by  an  accommodation  train  and  returns  by 
an  express  which  runs  10  miles  an  hour  faster.  .  Find  the  rates  of  the  two 
trains  in  miles  per  hour,  provided  that  the  time  occupied  for  the  two  trips 
was  one  hour  and  twenty-four  minutes. 

43.  It  is  found  that  two  steam  fire  engines  can,  by  working  together, 
pump  all  of  the  water  out  of  a  partly  filled  cellar  in  22^  minutes.  The 
more  powerful  one  alone  would  have  been  able  to  perform  the  work  in 
24  minutes  less  than  the  other  one  alone.  Find  the  time  required  by  each 
one  working  alone. 

44.  After  travelling  8  miles  in  an  automobile,  a  man  found  that,  on 
account  of  an  accident  to  the  machine,  it  was  necessary  to  walk  back.  If 
the  rate  of  the  automobile  exceeded  the  man's  rate  when  walking  by 
17  miles  an  hour,  and  he  was  2  hours,  16  minutes  longer  in  returning  than 
going,  find  the  rate  of  the  machine  in  miles  per  hour. 

Problems  in  Physics 

37.  If  a  moving  body  passes  over  equal  distances  in  successive 
equal  intervals  of  time,  the  motion  of  the  body  is  said  to  be  uniform. 
If  the  distances  passed  over  in  successive  equal  intervals  of  time 
are  not  equal,  the  motion  is  said  to  be  variable. 

38.  When  the  motion  of  a  body  is  uniform  its  vebcity  is  defined 
to  be  the  number  of  units  of  distance  passed  over  by  the  body  in 
one  unit  of  time.  When  the  motion  of  a  body  is  not  uniform  the 
velocity  at  any  instant  is  defined  to  be  the  number  of  units  of 
distance  which  would  be  passed  over  in  the  next  unit  of  time  if  the 
motion  of  the  body  were  to  become  uniform  at  that  instant. 

39.  If  the  velocity  of  a  moving  body  increases  during  successive 
intervals  of  time,  the  motion  is  said  to  be  accelerated. 

Acceleration  is  defined  to  be  the  rate  at  which  velocity  changes. 

Since  velocity  is  distance  per  unit  of  time,  it  follows  that  acceler- 
ation is  distance  per  unit  of  time  per  unit  of  time. 

Acceleration  is  said  to  be  positive  if  the  velocity  increases  in 
successive  intervals  of  time,  and  negative  if  the  velocity  decreases. 


510  FIRST  COURSE  IN  ALGEBRA 

E.  g.  A  body  which  is  falling  freely  from  any  point  above  the  surface 
of  the  earth  moves  toward  the  earth  with  uniformly  accelerated  motion. 
A  body  which  is  thrown  upward  moves  away  from  the  surface  of  the  earth 
^fnth.  a  uniformly  retarded  motion. 

40.  For  uniform  motion,  the  velocity,  v,  expressed  in  feet  per 
second,  of  a  body  which  in  t  seconds  passes  over  a  total  distance, 
St  expressed  in  feet,  may  be  found  from  the  formula 

S 

41.  Falling  Bodies.  If  a  body  in  a  state  of  rest  starts  to  fall 
from  any  point  above  the  surface  of  the  earth,  and  is  acted  upon  by 
the  force  of  gravity  alone,  the  total  distance  S^  expressed  in  feet, 
passed  over  in  t  seconds,  is  found  by  the  following  formula.  In 
this  formula  ^  is  a  numerical  constant  of  which  the  approximate 
value  may  be  taken  as  32  in  the  following  examples. 

s=hgt'- 

It  should  be  understood  that  the  results  obtained  by  using  the 
formulas  in  the  following  examples  are  only  approximate,  because 
the  value  32  substituted  for  g  is  approximate  and  because  no  al- 
lowance is  made  for  the  retarding  influence  due  to  the  resistance  of 
the  air. 

42.  The  velocity,  v,  expressed  in  feet  per  second,  of  a  falling 
body  at  the  end  of  t  seconds,  may  be  found  by  the  formula 

v  =  gt. 

Exercise  XXII.    11 
Solve  the  following  problems  relating  to  moving  bodies  : 

1.  What  velocity,  expressed  in  feet  per  second,  will  a  body  acquire  by 
falling  5  seconds  ? 

2.  In  what  time  will  a  falling  body  acquire  a  velocity  of  224  feet  per 
second  ? 

3.  What  is  the  height  of  a  tower,  if  a  stone  dropped  from  it  requires  3 
seconds  to  reach  the  ground  ? 

4.  A  stone  dropped  from  the  top  of  a  cliff  is  observed  to  reach  the  bottom 
in  5  seconds.     Find  the  height  of  the  cliff. 

5.  A  balloon  is  moving  horizontally  at  a  height  of  one  mile  above  the 
ground.     How  long  will  it  take  a  bag  of  ballast  to  reach  the  ground  ? 


PROBLEMS  IN  PHYSICS  511 

By  substituting  for  t  in  the  formula  S  =  -  gt^  the  value  -   ob- 

2  g 

tained  from   v  =  gt,   we  obtain 

v^  =  2gS. 

This  formula  may  be  used  to  find  the  velocity,  v,  in  feet  per 

second,  ac(|uired  by  a  body  falling  a  distance  of  S  feet. 

6.  What  velocity,  expressed  in  feet  per  second,  will  a  body  acquire  by 
falling  a  distance  of  576  feet? 

7.  What  velocity,  expressed  in  feet  per  second,  would  a  body  acquire  in 
falling  a  distance  of  500  feet  ? 

The  velocity  of  a  body  which  is  thrown  vertically  upward  with  a 
velocity  of  Vi  feet  per  second  will  be  retarded  by  an  amount  equal 
to  g  feet  per  second  per  second.  The  time  of  the  ascent  is  found 
by  dividing  the  initial  velocity,  Vi,  expressed  in  feet  per  second,  by 
the  constant  g* 

That  is,  ^  =  -'. 

9 

It  may  be  shown  that,  if  there  were  no  retarding  influence  due  to 
the  resistance  of  the  air,  the  times  required  by  the  body  in  ascend- 
ing and  descending  would  be  equal,  and  that  it  would  return  to  its 
starting  point  with  a  velocity  equal  to  that  with  which  it  was  thrown 
upward.  Hence  it  follows  that  the  height  to  which  a  body  will  rise 
when  thrown  vertically  upward  with  an  initial  velocity  of  v  feet  per 

second  is  given  by  S in  the  formula  S  =  —-- 

^9 

8.  A  stone  thrown  vertically  upward  strikes  the  ground  after  an  interval 
of  10  secon<ls.  With  what  velocity  was  it  thrown  and  to  what  height  did 
it  ri.se  ? 

9.  How  high  is  a  tree  if  it  requires  three  seconds  for  a  stone  which  is 
thrown  over  it  to  reach  the  ground  ? 

10.  With  what  velocity,*  expressed  in  feet  per  second,  must  an  arrow  be 
shot  vertically  upward  to  reach  the  top  of  a  tower  which  is  169  feet  high  ? 

If  a  falling  body  is  given  an  initial  velocity  downward  of  Vi  feet 
per  second,  the  total  space  S,  expressed  in  feet  per  second,  passed 
over  by  the  body  in  t  seconds,  may  be  found  by  the  formula 


612  FIRST  COURSE   IN   ALGEBRA 

11.  Find  the  distance  passed  over  in  three  seconds  by  a  body  which  is 
thrown  vertically  downward  with  a  velocity  of  24  feet  per  second. 

12.  Find  the  time,  expressed  in  seconds,  required  for  a  body  which  is 
thrown  vertically  downward  with  an  initial  velocity  of  5  feet  per  second 
to  move  a  distance  of  475  feet. 

13.  A  balloon  is  two  miles  from  the  ground  and  is  descending  at  the 
rate  of  eight  feet  per  second  when  a  sand  bag  is  dropped.  Find  the  number 
of  seconds  required  for  the  sand  bag  to  reach  the  ground. 

If  a  body  be  thrown  vertically  upward,  with  an  initial  velocity  of 
V  feet  per  second,  the  velocity  of  the  body  will  be  retarded  by  an 
amount  eiiual  to  g  feet  per  second  per  second. 

Accordingly,  since  the  acceleration  due  to  gravity  acts  in  such  a 
way  as  to  diminish  the  upward  motion  of  the  body,  it  follows  that 
the  initial  velocity  and  the  acceleration  due  to  gravity  must  be  con- 
sidered as  positive  and  negative  numbers. 

Hence,  if  a  body  is  projected  vertically  upward  with  an  initial 
velocity  of  ^i  feet  per  second,  the  velocity  Vt  at  the  end  of  t  seconds 
may  be  found  from  the  formula 

«\  =  ^h  —  gt. 
The  height,  expressed  in  feet,  to  which  the  body  will  rise,  is  repre- 
sented by  S  in  the  following  formula  : 

S=i\t-\gf. 

14.  A  balloon  is  500  feet  from  the  ground  and  ascending  at  the  rate  of  12 
feet  per  second  when  a  sand  bag  is  dropped.  How  many  seconds  will  be 
required  for  the  sand  bag  to  reach  the  ground  ? 

15.  If  a  body  is  projected  upward  with  an  initial  velocity  of  160  feet  per 
second,  what  is  the  height  to  which  it  will  rise  ? 

16.  A  rifle  bullet  is  shot  vertically  upward  with  an  initial  velocity  of 
400  feet  per  second.     Find  the  height  to  which  it  will  rise. 

17.  A  bullet  is  fired  vertically  upward  with  a  velocity  of  100  feet  per 
second.  Find  the  time  required  for  it  to  reach  a  point  156  feet  above  the 
ground,  and  also  the  velocity  with  which  it  passes  this  point. 


THEORY   OF  QUADRATIC    EQUATIONS  513 


CHAPTER  XXIII 

THEORY  OF  QUADRATIC  EQUATIONS  CONTAINING 
ONE  UNKNOWN 

1.  From  every  quadratic  equation  containing  one  unknown,  an 
equivalent  quadratic  equation  in  the  standard  form  a'3^  -{•  hx  -\-  c 
=  0  can  be  derived  by  making  suitable  transformations.  The  equa- 
tion aar*  +  bx-\-  c  =  0  has  been  shown  to  have  two  solutions.  Hence, 
it  follows  that  everi/  quadratic  equation  containing  one  unknown  has 
two  roots.     (See  Chapter  XXII.  §  23.) 

Denoting  the  two  solutions  of  the  standard  quadratic  equation 

aa^  +  bx  +  c  =  Ohyxi  and  x^,  we  may  write  Xi  = > 

2  a 


and  Xi  = 

2a 

In  all  of  the  discussions  which  follow,  it  will  be  assumed  that  a, 
b,  and  c  have  real  rational  values. 

2.  It  follows  from  the  Remainder  Theorem  (Chapter  VIII.)  that 
if,  when  any  value  represented  by  r  is  substituted  for  x  the  value 
of  the  quadratic  expression  aar*  -{-  bx  +  c  becomes  zero,  then  x  —  r 
is  a  factor  of  ax^  +  b^x  +  c. 

Since  ax^  -\-  bx  +  c  is  of  the  second  degree  with  reference  to  x, 
it  cannot  be  the  product  of  more  than  two  factors  which  are  each 
of  the  first  degree  with  reference  to  x. 

The  roots  of  the  quadratic  equation  ax^  +  bx  +  c  =  0  are  the 
roots  of  the  equations  obtained  by  equating  the  factors  of  the  first 
member  ax'^  +  bx  +  c  to  zero. 

Hence,  it  follows  that  the  quadratic  equation  ax^  +  bx  +  c  =  0 
cannot  have  more  than  two  roots.     (See  Chapter  XXII.  §  G.) 

33 


514  FIRST  COURSE   IN   ALGEBRA 

Nature  of  the  Roots 

3.  In  the  general  solution  x  — of  the  standard 

quadratic  equation  ax^  +  />a;  +  c  =  0,  the  expression  ^^  —  4  ac  is 
called  the  discriminant  of  the  quadratic  equation.  This  is  because 
it  affords  means  for  discovering  the  nature  of  the  roots  of  a  given 
equation,  —  that  is,  for  determining  whether  the  roots  are  posi- 
tive or  negative,  equal  or  unequal,  rational  or  irrational,  real  or 
imaginary. 

4.  When  values  represented  by  a,  6,  and  c,  in  the  standard  quad- 
ratic equation  a:i^  +  6ic  +  c  =  0,  are  selected  from  a  given  equation 
and  substituted  in  the  discriminant  h^  —  4  ac^  it  may  be  seen  that 
the  resulting  value  must  be  zero,  a  positive  number  or  a  negative 
number. 

By  referring  to  the  general  solution 

^_-h±  ^/b-  —  4:ac 

•C  —    

2a 

of  the  quadratic  equation  ax^  +  ^a;  +  c  =  0,  it  may  be  seen  that : 
(i.)    Ifb^—  4  ac  he  zero,  the  rcjots  are  real,  rational,  and  equal. 

That  is,        a^i  =  —  — .        and         ^  —  ~-^'  (See  §  1.) 

(ii.)  If  I?  —  A^  ache  positive,  the  roots  are  real  and  unequal,  and 
rational  or  irrationcd,  accm-ding  as  the  value  represented  hy  h"^  —  A^  ac 
is,  or  is  not,  the  squai'e  of  a  rational  number. 

(iii.)  If  h^  —  A:ac  he  negative,  the  expression  \/l/  —  4ac  repre- 
sents an  imaginary  quantity,  and  the  roots  are  conjugate  complex 
numbers. 

It  follows  from  the  general  solution  that  irrational  and  also  com- 
plex roots  enter  in  pairs. 

Determine,  without  solving,  the  nature  of  the  roots  of  each  of  the 
following  equations  : 

Ex.1.    Examine  x^  -\-Ax  +  A  =  Q. 

Substituting  in  the  discriminant  Ir^  —  Aac  the  values  from  the  given 
equation  represented  by  a,  b,  and  c  in  the  standard  quadratic  equation 
ax^  -f  6a;  +  c  =  0,  we  obtain 

42  -  4  •  4  =  0. 


THEORY  OF  QUADRATIC  EQUATIONS  515 

Since  the  value  represented  by  the  discriminant  is  zero,  it  follows  that 
the  roots  of  the  equation  are  real,  rational,  and  equal.  (See  (i.),  §  4,  and 
also  Chap.  XXII.  §  8,  Fig.  2.) 

Ex.  2.    Examine  ar^  +  4  a:  -  5  =  0. 

Substituting  in  the  discriminant  b^  —  4ac  the  values  from  the  given 
equation  represented  by  a,  6,  and  c  in  the  standard  quadratic  equation,  we 
obtain 

42  _  4.  i(_5)  =  36. 

Hence,  by  (ii.),  §  4,  the  roots  are  real  and  unequal.  Also,  since  the 
value  of  the  discriminant  is  the  square  of  a  rational  number,  —  that  is,  36 
is  the  square  of  the  rational  number  6,  —  the  roots  are  rational  and  un- 
equal.    (See  Chap.  XXII.  §  5,  Fig.  1.) 

Ex.  3.   Examine  a:^  -  8  a:  +  1 1  =  0. 

Substituting  in  the  discriminant  b^  —  4:ac  the  values  from  the  given 
equation  represented  by  a,  6,  and  c  in  the  standard  quadratic  equation,  we 
have  (-8)2-4.1.11  =  20. 

Since  the  value  represented  by  the  discriminant  is  a  positive  number, 
20,  it  follows  from  (ii.),  §  4,  that  the  roots  are  real  and  unequal. 

Also,  since  20  is  not  the  square  of  a  rational  number,  that  is,  ^/^  is 
irrational,  the  roots  are  conjug.ite  irrational  numbers. 

Ex.  4.   Examine  a;^  +  4  a:  +  5  =  0. 

The  value  represented  by  b^  —  4ac  is  negative,  that  is, 
42  _  4  .  1  .  5  =  _  4. 

Accordingly,  by  (iii.),  §  4,  the  roots  are  imaginary  ;  that  is,  they  are 
conjugate  complex  numbers.     (See  Chap.  XXII,  §  9,  Fig.  3.) 

Ex.  5.   Examine        mx^  -}-  (m  ■}-  n)  x  -\-  n  =  0,  m  7^  n. 

Comparing  with  the  standard  quadratic  equation  ax^  -{-  bx  -{■  c  =  0,  we 
find  that  m  is  represented  by  a,  (m  +  n)  by  &,  and  nhy  c. 

Hence,  substituting  these  values  in  the  discriminant  b^  —  Aac,  we  obtain 

(m  +  n)2  —  4  mn  =  (m  —  n)^. 

Since  the  square  of  any  real  number  is  positive,  it  follows  from  (ii.),  §  4, 
that  the  roots  are  real,  rational,  and  unequal,  for  real  values  of  m  and  n. 

Exercise  XXIII.     1 
Determine,  without  solving  the  equations,  the  nature  of  the  roots 
of  the  following : 

1.  x^  +  10a;  +  25  =0.  3.  2a?'  +  ^x  —  27  =  0. 

2.  9£c2- 24ic+  16  =  0.  4.  Gx'-x-  15  =  0. 


516  FIRST  COURSE   IN   ALGEBRA 

5.  3a^- 5;r  +  1  =  0.  12.  x^  +  Ax-  11  =0. 

6.  a;^+ 6ic— 247  =  0.  13.  3ic^  +  2a;  +  2  =  0. 

7.  3ic  =  ar^  +  4.  14.  2  +  9a;  -  Sic^  =  0. 

8.  a;2+ 3a;=  180.  15.  4ar' -  5a;  +  3  =  0. 

9.  7 a;^  —  8a;  +  2  =  0.  16.  ar»  —  a;  —  342  =  0. 

10.  4ar'— 10a;+  3  =  0.  17.  6ar^  +  5a;  +  4  =  0. 

11.  a;^  — a;H- 1  =0.  18.  ar*  —  a;  -  756  =  0. 

19.  Prove  that  the  roots  of  a;^  —  2  aa;  +  a^  =  ^^  +  c^  are  real. 

20.  Prove  that  the  roots  of  2  aar^  +  (2  «  +  3  ^>)  a;  +  3  6  =  0  a,re 
real  for  all  real  values  of  a  and  b. 

21.  Prove  that  3  cx^  —  (2  c  +  3  d)x  +  2  of  =  0  has  rational  roots. 

22.  Show  that  the  roots  of  5  a;^  +  4  aa;  +  fl'^  =  0  are  imaginary. 

23.  Show  that  the  roots  of  {a  +  A) V^  -  2(«2  —  b'^)x  +  {a  -  bf  =  0 
are  neither  irrational  nor  imaginary. 

Determine  the  value  which  k  must  have  in  order  that  the  following 
equations  shall  have  equal  roots  : 

Ex.  24.  x^  +  (k  -  3)a;  +  ^'  =  0. 

The  condition  for  equal  roots  is  that  the  discriminant  (k  —  3y  —  Ak  shall 
be  zero.     (See  (i.),  §  4.) 

Accordingly,  placing  the  discriminant  equal  to  zero,  we  obtain  the  con- 
ditional equation, 

(^- _  3)2  _  4^-0, 

the  solutions  of  which  are  A:  =  9,  and  k  =  I.     (See  (i.)  §  4.) 

It  will  be  found,  if  9  is  substituted  for  k,  that  the  given  equation  will 

reduce  to  x^  +  6  a:  +  9  =  0,  which  has  equal  roots. 

Also,  substituting  1  for  k,  the  given  equation  reduces  to  a:^  —  2  a:  +  1  —  0, 

which  also  has  equal  roots. 

25.  a;2  =  2A:(a;  — 4)  +  15.  32.   12^a;2  —  2a;  +  3^  =  0. 

26.  x"  +  2(k  +  2)a;  +  9^  =  0.  33.  a;^  -  (2^  —  3)a;  +  2^  =  0. 

27.  ar'  — 2^a;+  6a;+  4^^  =  0.  34.  (8^  +  5)ar^  — 12^a;+ 1  =  0. 

28.  it'  +  k(2x-S)  =  15.  35.  (^4-2)a;H3^+4=5(^-l)a% 

29.  a^  =  (k  —  l)x  -  2(k  —  1).  36.  ar'+(6/;+7)a;+6^+22=0. 

30.  (2^+l)a;2  +  3^a;  +  ^  =  0.  37.  (^+ 6)a;'-3(^-2)a;=l-A:. 

31.  ar*  +  Skx  +  4^+1  =  0.  38.  (^-ll)a;2+3^+4=2(^-l)a;. 


THEORY   OF   QUADRATIC   EQUATIONS  517 

Relations  Between  Roots  and  Coefficients 

5.   Representing  the  two  roots  of  the  standard  quadratic  equation 
ax^  +  ^  +  c  =  0  by 

—  h^-  Vb^  —  4:ac        ,  -h—  ^/h^  —  ^ae 

Xi  = i  ana  X2  = ; > 

2  a  2  a 

as  in  §  1,  above,  it  follows  immediately  by  addition  that 


[-  b  4-  Vb^  -Aac]  +  [-b-  Vb^-4.ac] 
x,  +  x,=  —  , 

or     ici  +  iCa  = (1) 

ci 

Writing  the  product  of  the  roots,  we  have 

^^^ = L — ^ — J  L — ^ — J 

_  4-  4ac 

Hence 

c 
xix^  =  --  (2) 

d 

If  the  terms  of  a  quadratic  equation  in  standard  form, 
ax^  +  6ic  +  c  =  0, 
be  all  divided  by  «,  which  is  the  coefficient  of  £c^,  the  coefficient  of 
x^  will  be  unity  in  the  derived  equivalent  equation 

aj2  +  -  a;  +  -  =  0.  (3) 

a         a  ^ 

Referring  to  (1)  and  (2),  it  may  be  seen  that  in  (3)  : 

(i.)    The  sum  of  the  roots  xi  and  x^  is  equal  to  the  coefficient  of 

X  with  its  sign  changed 

(ii.)   The  product  of  the  roots  is  equal  to  the  term  free  from  x, 

,       •      c 
that  IS,   -  • 
a 

Principles  (i.)  and  (ii.)  may  be  used  as  checks  upon  the  solution 
of  a  quadratic  equation. 


518  FIRST  COURSE   IN  ALGEBRA 

E.  g.    The  roots  of  2  x^  +  7  x  +  3  =  0  are  -  3   aiul   -1/2. 

Dividing  the  terms  by  2,  which  is  the  coefficient  of  x^,  we  obtain  the 

7         3 
equivalent  equation,  a:*  +  -a5  +  -  =  0. 

It  may  be  seen  that  the  sum  of  the  roots  —  3  and  —  1  /2  is  the  coefficient 
of  X  with  its  sign  changed,  that  is,  —  ^,  and  the  product  of  the  roots  is  ^. 

c 

6.  From  xix^  =  - ,   (2),  obtained  in  §  5,  it  appears,  if  the  roots 

Xi  and  a?2  are  real,  that  according  as  the  numbers  represented  in 
the  standard  equation  by  <i,  and  c  have  like  or  unlike  signs,  the 

quotient  -  will  be  positive  or  negative,  and  the  roots  of  the  given 

u 

equation  will  be  both  positive  or  both  negative. 

E.  g.   CJonsider  2  re*  _  7  x  +  3  =  0. 

Using  the  discriminant  we  find  tliat  the  roots  cannot  be  imaginary,  since 
(—  7)^  —  4  •  2  •  3  =  25,    which  is  a  positive  number. 

The  quotient  f ,  represented  by  - 1  is  positive.     Hence  the  roots  cannot 

differ  in  sign,  and  must  be  either  both  positive  or  both  negative. 

It  may  also  be  seen  that  the  roots  of  the  equation  2x=^  +  5x  —  3  =  0  are 

also  both  real,  but  they  have  opposite  signs  because  —^  is  negative. 

7.  From  a*i  +  3*2  = >   (1)}  §  5,  it  appears  that  the  sum  of  the 

roots  is  represented  by Hence  the  roots  either  both  agree  in 

sign  with  the  quotient  -  with  its  sign  changed^  or  if  they  differ  in 
sign,  the  sign  of  the  greater  root  must  agree  with  the  reversed  sign  of 
the  qtiotientj  -  •     (See  (i.)  §  5.) 

From  this  it  follows  that,  if  the  number  represented  by  a  in  the 
standard  quadratic  equation  aa:^  +  bx  -\-  c=^0  is  positive,  the  root 
which  is  numerically  the  greater  is  opposite  in  sign  to  the  sign  of  the 
number  represented  by  b. 

E.  g.  The  roots  of  2x2  —  5x  —  3  =  0  ^re  both  real,  but  they  have  oppo- 

—  3   . 

Bite  signs,  since  -^-  is  negative. 


THEORY   OF   QUADRATIC   EQUATIONS 


519 


The  greater  root  must  be  positive  because  its  sign  must  agree  with  the 
reversed  sign  of  the  quotient  —  ^. 

For  convenience  of  reference,  the  illustrations  used  above  are  given  in 
tabular  form  as  follows  : 


Equation. 

c 
a 

The  roots 
are 

6 
a 

Signs  of 
roots  are 

Greater 
root 

1^- 
1'- 

7    . 

-7  . 
2    " 

Both  - 
Both  + 

+ 

.2x2  +  5x-3  =  0 
2a:2-5x-3  =  0 

-3. 

-3. 

2    '^ 

2  '^ 

Different 
Different 

+ 

Formation  of  an  equation  liavinj?  Specified  Roots 

8.  A  quadratic  equation  having  specified  roots  may  be  con- 
structed by  applying  Principles  (i.)  and  (ii.)  §  5. 

—  b  c . 

We  may  substitute  xi  -f-  x^  for and  Xix^  for  -  in  the  quadratic 

equation  x^  -\ — x  •{ —  =  0  and  obtain 
a        a 

X^  +  [—  (Xi  +  X<i)]x  +  [XiXi]  =  0. 

A  quadratic  equation  having  specified  roots  may  be  obtained  by 
constructing  a  quadratic  equation  of  which  the  coefficient  of  x^  is 
unity,  the  coefficient  of  x  is  the  sum  of  the  roots  with  sign  changed, 
and  the  term  free  from  x  is  the  product  of  the  specified  roots. 

Whenever  fi:actions  appear  in  any  of  the  terms  of  an  equation 
thus  constructed  an  equivalent  integral  equation  may  be  obtained 
by  multiplying  the  terms  by  the  lowest  common  multiple  of  all  the 
denominators  of  the  firactions. 

Ex,  1.    Construct  the  quadratic  equation  the  roots  of  which  are  5  and  7. 
The  sum  of  the  roots  is  5  +  7  =  12,  and  the  product  is  5  •  7  =  35. 
Hence,  changing  the  sign  of  the  sum  12,  we  may  write  as  the  required 
equation,  a^  —  12  x  +  35  =  Q. 


520  FIRST  COURSE   IN  ALGEBRA 

Ex.  2.     Construct  the  equation  the  roots  of  which  are  +  3  anil  —  ^. 
The  sum  of  the  roots  is  +  S  —  8  =  —  5,  and  the  product  is  3(—  8)  =  —  24. 
Changing  the  sign  of  the  sum,  —  5,  the  equation  required  is 

x2  +  5  X  -  24  =  0. 
Ex.  3.     Constnict  the  equation  the  roots  of  which  are  J  and  —  f . 
The  sum  of  the  roots  is  4  ~  f  =  ?V 

The  product  of  the  roots  is         (i)(—  §)  =  —f^' 
Changing  the  sign  of  the  sum,  we  may  construct  the  equation 

Or,  24x2-5  a; -14  =  0. 

Ex.  4.     Construct  the  equation  the  roots  of  which  are  —  and  —  1. 

The  sum  of  the  roots  is  1  = 

n  n 

The  product  of  the  roots  is  (  —  )(—  M  = * 

Using  the  sum,  with  its  sign  changed,  as  the  coefficient  of  x,  we  have, 

x^ X =  0. 

n  n 

Or,  iix"^  —  (m  —  n)x  —  m  =  0. 

Ex.  5.   Construct  the  equation    the  roots  of  which  are  the  conjugate 

irrational  numbers  2  +  \/5  and  2  —  ^b. 

The  sum  of  these  values  is  (2  +  ^5)  +  (2  -  V^)  =  4. 
The  product  is  (2  +  V5)(2  -  ^/b)  =  4  -  5  =  -  1. 

Reversiug  the  sign  of  the  sum,  we  may  write  the  equation 
x2_4x- 1  =  0. 

Ex.  6.    Construct  the  equation  the  roots  of  which   are   the   complex 

_  1  +  ^^32                 -  1  -  V^ 
numbers  tt^ and      -^^ 

_  1  +  V-^       —  1  —  a/— ~2       —  2 
The  sum  of  these  values  is     ^ 1 ^-^^ =  -^  • 

The  product  is, 

Using  the  sum  with  its  sign  changed,  we  have  for  the  equation 

a;2+  2^4-^  =  0. 

Or,  3x24-2a:  + 1  =0. 


THEORY  OF  QUADRATIC  EQUATIONS       521 

9.  The  roots  of  a  quadratic  equation  in  the  standard  form 
ax^  +  6.r  +  c  =  0  are  the  roots  of  the  two  linear  eqiuitions  fm^med  by 
equating  to  zero  the  two  linear  factms  the  product  of  which  is  the 
first  member  of  the  equation.     (See  Chap.  XII.  §  48.) 

It  follows  that,  by  reversing  the  process,  we  may  construct  a 
quadratic  equation  having  specified  roots  by  equating  to  zero  the 
product  of  the  two  linear  factors  which^  when  equated'  to  zero  and 
considered  as  equations,  have  as  roots  the  given  values. 

Ex  7.    Construct  the  equation  the  roots  of  which  are  5  and  7. 

We  may  indicate  that  these  are  roots  x^  and  z^  of  an  equation  by  writing 
aTj  =  5     and    x^  =  7. 

Accordingly,  x^  —  5  =  0,     and    Xg  —  7  =  0. 

These  two  equations  taken  together  are  equivalent  to  the  single  quadratic 
e(iuation  (x^  -  5)(x2  —  7)  =  0. 

Performing  the  indicated  multiplication  and  neglecting  subscripts,  we 
obtain  x'^  -  12x  +  35  =  0.     (Compare  with  Ex.  1,  §  8.) 

10.  It  will  be  found  that,  whenever  the  roots  are  irrational  or 
imaginary,  the  method  of  §  8  is  to  be  preferred. 

Exercise  XXIII.     2 
Construct  integral  equations  the  roots  of  which  are 

1.  3  and  5.  15.  —  ^  and  t^. 

2.  2  and  9.  16.  —  ^  and  2. 

3.  G  and  8.  ^17.  f  and  3. 

4.  4  and  7.  18.  5/a  and  -  ^/6. 

5.  I  and  2.  Id.  m/n  and  n/m. 
6.-5  and  —  8.  20.  b/2c  and  c/2  b. 
7.-7  and  —  10.  a  —  b 

8.  -3  and +  5.  21.  ^-;^  and  +  1. 

9.  +  2  and  —  15. 

10.  2  and  -2.  22.  t-?-  and - 


11.  0and4. 


b  +  c 


12.  f  and  f  23.  2\/5  and  —  2^5. 

13.  ^  and  i.  24.  Vs  and  -  a/3. 

14.  f  and  §.  25.  1  +  V^  and  T  —  Vg 


522  FIRST   COURSE   IN   ALGEBRA 

26.  2  +  A/^^and  2  —  a/^.     28.  c+  V^=*— 1  andc— Vc^-l. 

27.  «+V?ianda-V6.  29.  ?±^ and ^i:^ . 

()  6 

11.*  One  Root  Known.  When  one  root  of  a  given  quadratic 
equation  is  known,  by  applying  principle  (i.),  §  5,  the  other  root 
may  be  found  immediately  without  solving  the  equation. 

Ex.  1.    Knowing  that  one  root  of  x^  +  5x  —  24  =  0  is  3,  find  the  other. 

Since  the  coefficient,  5,  of  x  with  iUs  sign  changed  is  tlie  sum  of  the  roots, 
we  may  obtain  the  required  root,  —  8,  by  subtracting  the  given  root  3  from 
-  5,  that  is,  —  5  -  3  =  -  8. 

The  root  may  also  be  obtained  by  dividing  the  known  term  —  24  by  3. 
(Compare  with  Ex.  2  §  8.) 

Ex.2.  Knowing  that  one  root  of  the  ecpuition  a;^  — 4a;— 1=0  is 
2  —  /y/S,  we  can  find  the  remaining  root  by  subtracting  2  —  \/5  from  the 
coefficient  of  x  with  its  sign  changed. 

We  have,  +  4  -  (2  -  V^)  =  2  +  ^5.     (Compare  with  Ex.  5,  §  8.) 
Since  in  an  efjuation  in  which  the  coefficients  are  rational,  irrational  or 
complex  roots  enter  in  conjugate  pairs,  if  indeed  they  enter  at  all,  it  follows 
that  we  may  obtain  the  reipiired  root  immediately  by  writing  the  binomial 

2  +  -V^,  which  is  the  conjugate  of  the  one  given,  2  —  ^E. 
Exercise  XXIII.     3 

(ThiB  exercise  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

Find,  without  solving,  the  remaining  root  of  each  of  the  following 
equations,  when  one  of  the  roots  is  given  : 

1.  ic^  —  3  a;  —  130  =  0,  one  root  being  13. 

2.  ic*  +  2  a;  —  323  =  0,  one  root  being  —  19. 

3.  Qc^  —  21x—  130  =  0,  one  root  being  —  5. 

4.  4  ar*  4-  28  aj  —  15  =  0,  one  root  being  ^. 

5.  18  ar^  ~  117  a;  +  37  =  0,  one  root  being  ^. 

6.  25  a^  —  85  a;  —  18  =  0,  one  root  being  —  ^. 

7.  3  ar*  —  5  a;  —  308  =  0,  one  root  being  11. 

8.  98  a;'^  —  7  a;  —  6  =  0,  one  root  being  f . 

9.  0^  —  (a  +  b)x+  2  ah  —  2  ^^  =  0,  one  root  being  a  —  b.^ 

*  This  section  may  be  omitted  when  the  ciiapter  is  read  for  the  first  time. 


REVIEW  523 

10.  2  ax^  +  (4  —  ah)x  =  2b,  one  root  being  b/2. 

11.  cdx^  —  {(•  -r  d)x  +1=0,  one  root  being  1/c. 

12.  6  aV  +  5  «ic  +  1  =  0,  one  root  being  —  1  /2  a. 

13.  4:X^  —  Aax  =  b^  —  a^,  one  root  being  (a  -\-  b)/2. 

14.  ab^cx^  —  b(c  —  a)x  —1=0,  one  root  being  1/ab. 

15.  4  abx^  +  2  (a^  +  b'^)x  +  ab  =  Oy  one  root  being  —  «/  2  6. 

16.  acx^  -j-  b(a^  +  c^)x  +  ab^c  =  0,  one  root  being  —  abj  c. 

Mental  Exercise.     XXIII.     4.     Review 
Cube  each  of  the  following : 


1.  x. 

5.  -  b\ 

9.  x^. 

13.  a-K 

2.  y\ 

.-'.. 

10.  ijK 

14.  ri 

3.  z\ 

11.  zl 

15.  c~K 

4.  -a.  8.  -f.  12.  ^-  16.  (TI 

Express  each  of  the  following  numbers  as  a  power  of  2  : 

17.  (4«)l  18.  (2«)^  19.  (16^)^  20.  (32')». 

Solve  each  of  the  following  equations  : 

21.^^=1.  22.25/=l.      2S.'-^=i.       24.g  =  4 

Factor  each  of  the  following  expressions : 

25.  a^  -  10  ab  +  25  b""  —  36.         27.  9  w^  +  23  twV  +  16  n\ 

26.  2^^+180;^  +  45/.  28.  15aj''  +  13a;^  +  2/. 

Solve  each  of  the  following  equations: 


29.  x^  =  3. 

32.  w^  =  -2. 

35.^/ =  3. 

30.  y^  =  2. 

33.  x^  =  -l. 

36.  x-^  =  -  r' 

31.  z^  =  i 

34.V/|  =  3. 

37.  'v/S  =  w. 

Express  the  following 

as  positive  fractions  : 

38.           ', 

39.-*:^ 

40.  ^-^^T"- 

1  —  a  3  ic  + 


524  FIRST  COURSE  IN  ALGEBRA 

Express  the  following  without  negative  exponents: 


41.  (-a)-l 

42.   (-  b)-\ 

43. 

(-O-s. 

Simplify  each  of  the  following  expressions : 

44.  -2aa-\ 

46.  aia^y. 

48. 

(2«-r6V. 

«s- 

49. 
50. 

Distinguish  between 

— -            1 
51.  a   *  and   -i* 
a* 

52.  -  a' 

and 

-i 
a  ^- 

53.  Express  (  y  j^  as  a  quotient  of  powers. 

Find  the  value  of 

54.  .2-^  55.  .04-\  56.  .5"^  57.  (.02)" 

Rationalize  the  denominators  of  each  of  the  following  : 
58.  -4=-  60.  -^-  62.      ^ 


V2  '  ^  '  \/bb 

12  2 

59.  -^^'  61.  — =.  63. 


V^4  V2  a  \/2c 

What  roots  may  possibly  be  introduced  by  squaring  both  members 
of  each  of  the  following  equations  ? 

64.  jc  =  5.  66.  ;?  -  1  =  6.  68.  ^  —  3  =  7. 

65.  :j/  +  2  =  3.  67.  a  —  2  =  0.  69.  ;^  +  8  =  3. 

Distinguish  between 

70.  (-  a/2)'  and  W^^f-  71.  -  ^8  and  ^^^. 

72.  V6  —  V6  and  V6  +  V'^- 

Simplify  each  of  the  following  expressions  : 

73.  (V5  +\/=^)^      74.  (V=l  -  V^y.     75.  (2a^  +  i)'. 

Solve  each  of  the  following  equations  : 

76.  0  =  2^/7  Vic  -  7.  77.  a/«  -  «  =  0. 

78.  (a;  -  3)(9  •  7  •  i  +  32  •  16  -  5)  =  0. 


REVIEW  525 

Show  that  the  following  identities  are  true  : 
19.—i  =  \-  SO.  —i  =  i-\  Sl.—i-^=l. 

82.  (a-b)(c -b){d- c){d -a)  =  (a- b)(b - c)(c - dXd-a). 

83.  Rationalize  — -  •  84.  Realize 


V2  a/-2 

85.  Find  the  values  of  {a'  +  by  and  (a'  +  by. 
Simplify  each  of  the  following  : 

86.  -  (a-i^/-V«)-^.       87.  -  {x-^ -^  y-y\     88.  (wv^)". 


526  FIRST  COURSE  IN   ALGEBRA 


CHAPTER  XXIV 

IRRATIONAL  EQUATIONS  AND  SPECIAL  EQUATIONS 
CONTAINING  A  SINGLE   UNKNOWN 

Irrational  Equations 

1.   An  irrational  equation  is  an  equation  in  which  one  or  more 
terms  are  irrational  with  reference  to  the  unknown. 

E.g.  Vx^-S  =  x-l. 


V^ar  +  S  -I-  y'x-f.  12  =  7. 


^x  +  4  -  2  Va:  +  13  +  Vx  +  22  =  0. 

2.  It  is  understood  that,  when  the  terms  of  an  equation  are  af- 
fected by  radical  signs,  principal  values  only  of  the  roots  are  to  be 
taken,  unless  we  have  means  for  knowing  that  other  values  should 
be  t^ken.  (See  Chap.  XVIII.  §  13,  also  Chap.  XXIV.  Ex.  3,  §  6,  and 
Ex.  5  and  Ex.   6,  §  10.) 

3.  To  be  consistent  it  is  necessary  that  a  specified  letter  represent 
the  same  numerical  value  wherever  it  appears  in  the  terms  of  a  given 
conditional  equation. 

E.  g.  Ill  the  equation  a:^  —  5a:  +  fj  =  0,  x  may  represent  either  3  in 
every  term  or  2  in  every  terra,  but  never  3  in  one  term  and  2  in  another 
term. 

Thus  the  equation  x^  —  5a;  +  6  =  0  represents  either  of  the  following 
numerical  identities  : 

32  _  5  .  3  +  6  =  0     or     22  -  5  •  2  +  6  =  0. 

It  is  also  necessary  that  a  specified  symbol  of  operation,  such  as 
\'^j  represent  the  same  root  wherever  it  may  appear  among  the 
different  terms  of  an  equation. 

It  may  be  seen  that  if  V  is  to  be  considered  as  representing  the 
positive  value  of  the  root  in  one  term,  to  be  consistent  we  must 


IRRATIONAL  EQUATIONS  527 

understand  it  as  representing  the  positive  value  of  the  root  in  every 
other  term  of  the  equation. 

Thus,  V  cannot  represent  a  positive  value  of  the  root  in  one 
term  of  an  equation  and  a  negative  value  in  another  term. 

4.  When  solving  irrational  equations  it  is  necessary  to  consider 
the  following 

Principle  Relating  to  Extra  Roots:  Whenever  both  members 
of  an  equation  are  raised  to  the  same  positive  integral  power  no  solu- 
tions of  the  original  eqmition  are  lost  in  the  process ;  but  solutions 
may  be  gained  which  satisfy  the  derived  equation^  yet  do  not  satisfy 
the  original  equation. 

Let  the  members  of  a  given  equation  be  represented  by 

A=B.  (1) 

Raising  both  members  to  the  nth.  power,  n  being  a  positive  in- 
teger, we  obtain  A''  =  /?".  (2) 

Transposing,  ^"  -  7i"  =  0.  (3) 

For  all  positive  integral  values  of  w,  ^"  —  ^"  is  exactly  divisible  by 
A  —  By  and  the  quotient  Q  resulting  from  the  division  is  of  degree 
n-\. 

Accordingly,  factoring  ^"  —  Z?",  we  may  from  (3)  obtain 

{A-B)Q  =  ^'  (4) 

This  last  equation  (4)  is  equivalent  to  the  set  of  two  equations 
^  -  ^  =  0,  •  (5)    and     Q  =  0.     (6) 

Equation  (r>)  is  equivalent  to  the  original  equation  A=  B  (1), 
and  ((i)  is  an  additional  equation  introduced  by  the  process  of  rais- 
ing both  members  of  (1)  to  the  nth.  power. 

Since  equation  (2)  and  its  equivalent  equation  (4)  are  satisfied 
by  any  solution  either  of  equation  (5)  or  of  equation  (6),  it  follows 
that  any  solution  of  the  additional  equation  Q  =  0  (6),  which  is 
not  at  the  same  time  a  solution  of  equation  (5)  must  have  been 
introduced  by  the  process  of  raising  the  members  of  (1)  to  the  nth. 
power. 

E.  f?.    If  both  members  of  .r  +  2  =  5  be  raised  to  the  second  power,  we 
shall  obtain  the  equation  {x  +  2)2  =  25,  which  has  two  solutions, 
ar  =  3     and     x  =  —  l. 


528  FIRST  COURSE  IN  ALGEBRA 

The  first  of  these  values,  3,  is  the  single  solution  of  the  given  equation, 
while  the  second  value,  —  7,  clbes  not  satisfy  the  given  equation,  but  was 
introduced  by  the  process  of  squaring. 

5.  Irrational  equations  are  ustialli/  prepared  for  solution  as 
follows  : 

First  the  terms  are  so  transposed  that  one  member  of  the  equation 
consists  of  one  of  the  irrational  terms  appearing  in  the  equation^  all 
other  terms  being  transposed  to  the  remaining  member. 

Both  members  of  the  equation  are  then  raised  to  the  lowest  power 
necessary  to  rationalize  the  term  which  has  been  separated  from  the 
others. 

If  irrational  terms  still  appear  after  the  operation^  the  process  is 
repeated  until  an  equation  is  obtained  which  is  entirely  rational  with 
respect  to  the  unknown  quantity  appearing  in  it. 

All  solutions  of  the  rational  equation  thus  obtained  which  do  not 
satisfy  the  given  irrational  equation  must  be  rejected  as  being  extra 
solutions  introduced  during  the  process  of  rationalization. 


Ex.  1.    Solve                                       14  -  a:  =  ^\m  -  x\ 

(1) 

Squaring  both  members,     196  -  28  x  +  x^  =  100  -  a:^. 

(2) 

The  equivalent  equation       a:^— 14x  +  48  =  0 

(3) 

has  the  two  solutions,  x  =  8    and    x  =  6. 

We  find  by  substitution  that  both  values  satisfy  the  given  equation. 
Hence  in  this  case  no  solutions  have  been  introduced  by  the  process  of 
squaring  the  meml>ers  of  the  given  equation. 

If  the  given  equation  (1)  be  written  in  the  form 


14-x- /v/l00-x2  =  0,  (4) 

it  may  be  seen  that  a  rational  equation  can  be  derived  from  it  by  using  the 
rationalizing  factor  14  —  x  +  >v/lOO  —  x^. 

Any  root  which  may  possibly  be  introduced  into  the  derived  e(|uation 
by  the  use  of  this  factor  must  be  a  root  of  the  equation  obtained  by  placing 
this  factor  equal  to  zero,  —  that  is,  of  the  equation 


14  _  a;  +  yiOO  -  x2  =  0, 


or,  14  -  X  =  -  ^/\00  -  x\  (5) 

Either  of  the  solutions  of  equation  (3),  x  =  8  or  x  =  6,  if  substituted  in 
(5),  makes  the  first  member  positive  and  the  second  member  negative,  pro- 
vided that  we  take  only  the  principal  value  of  the  root. 


IREATIONAL  EQUATIONS  529 

Hence,  with  the  restriction  that  the  principal  value  only  of  the  root  is  to 
be  taken,  these  values  could  not  have  been  introduced  during  the  process  of 
rationalization. 

6.  It  should  be  observed  that  we  may  write  an  equality  which 
expresses  a  relation  between  two  numbers  or  expressions  which  is 
inconsistent  with  the  laws  governing  relations  between  numbers. 

Such  an  equality  is  sometimes  spoken  of  as  an  "impossible 
equation." 

E.  g.  Thus,  since  it  is  impossible  that  a  positive  number  be  equal  to  a 
negative  number,  equation  (5)  of  §  5  is  an  illustration  of  an  "inqjossible 
equation." 

Rational,  fractional  equations  having  no  finite  solutions  may  be 
written. 

E.  g. = expresses  a  condition  which  cannot  be  satisfied 

X  —  2       X  —  3 

by  any  finite  value  of  a;,  that  is,  it  is  a  so-called  "impossible  equation." 

It  may  be  seen  that  by  clearing  of  fractions  we  obtain  the  inconsistent 

relation   a:  —  2  =  a:  —  3,   or  — 2  =  —  3,   which  is  impossible. 


Ex.  2.    Solve  a:  -  5  +  y'a;  -  5  =  0.  (1) 

Transposing,  ^x  —  5  =  5  —  a:.  (2) 

Squaring  both  members,  a:  —  5  =  25  —  10  a:  +  a;^,  ^3^ 

Equation  (3)  is  equivalent  to  a;^  —  11  a:  +  30  =  0,  (4) 

of  which  the  solutions  are  x  =  6,   and  x  =  5. 

Both  values  are  roots  of  the  rational  equations  (3)  and  (4),  but  if  we 
restrict  the  roots  to  principal  values  only,  equation  (I)  is  satisfied  by  the 
value  X  =  5,   but  not  by  the  value  x  =  6. 

It  appears  that  x  =  6  is  an  extra  root  introduced  by  squaring  the 
members  of  (2). 

It  should  be  observed  that  if,  instead  of  first  transposing  the  terms  of 
equation  (1)  and  then  squaring,  we  had  obtained  a  rational  equation  by 
multiplying  both  members  of  (1)  by  the  rationaliziiig  factor  x  —  5  —  y\/x  —  5, 
we  would  have  obtained  the  same  rational  equatipn  (4)  as  before,  and  con- 
sequently the  same  solutions,   x  =  6,    and   x  =  5. 

We  find  that  both  of  these  values  satisfy  the  "multiplier  equation," 
x  —  5  —  -y/x  —  5  =  0,  formed  by  equating  the  rationalizing  factor  to  zero. 
Hence,  either  of  the  values  a:  =  6  or  a:  =  5  might  have  been  introduced 
during  the  process  of  rationalization. 

It  is,  therefore,  necessary  to  substitute  in  the  original  equation  to  find 

34 


530  FIRST  COURSE  IN   ALGEBRA 

whether  or  not  either  or  both  of  these  values  can  be  accepted  as  solutions 
of  the  given  equation  (1). 

Ex.  3.    Solve  X  +  ^y/x  —  2  =  0. 

It  should  be  observed  that  this  equation  is  quadratic  with  reference  to 
-y/x.     Hence  we  may  factor  with  reference  to  /y/a:,  and  obtain, 

(V^+2)(v^- 1)  =  0. 
Hence,  ^/x  +  2  =  0,  and  /y/x  —1=0. 

Or,  ^x  =  —  2,        and  ^x  —  1. 

Hence,  a:  =  4,  and  x  =  1. 

It  should  be  observed  that  tlie  value  4  is  the  square  of  the  nef^ative 
number  —  2.  Hence,  when  substituting  4  for  x  in  the  given  equation,  it 
is  necessary  that  ^x  be  considered  equal  to  —  2. 

With  this  understanding,  it  will  be  found  that  the  value  4  satisfies  the 
given  equation. 

For,  we  have  4  +  (-  2)  -  2  =  0.     That  is,  0  =  0. 

The  remaining  value  x—\  will  be  found  by  substitution  to  satisfy  the 
given  eci nation. 

For,  we  have  1  +  1-2  =  0.     That  is,  0  =  0. 

7.  If  ^,  5,  and  C  be  any  functions  of  x^  it  may  be  shown  that 
the  equations 

\/J+ \/5+ v^  =  o,       (1) 

VI +^^-^^^=0,  (2) 

VZ-V^+VC'=0,         (3) 

V3"-v^-Va  =  o,       (4) 

when  rationalized,  all  lead  to  the  same  derived  equation 

A^'-^B'^C-^AB-^BC-'iGA^^.  (5) 

Accordingly,  equation  (5)  is  equivalent  to  the  set  of  four  equations 
(1),  (2),  (3),  and  (4). 

It  follows  that,  when  solving  an  irrational  equation  having  the 
form  of  any  one  of  the  irrational  equations  (1),  (2),  (3),  or  (4),  we 
shall  obtain  the  solutions,  not  only  of  the  given  equation,  but  also 
of  the  equations  represented  by  the  remaining  three  equations  of  the 
set,  obtained  by  changing  among  themselves  in  all  possible  ways  the 
signs  of  the  radicals  of  the  given  equation. 


IRRATIONAL  EQUATIONS  531 

8.  We  cannot  speak  of  the  degree  of  an  equation  which  is  irra- 
tional with  reference  to  the  unknown,  and  we  have  no  means  for 
knowing  before  solving,  as  we  have  in  the  case  of  integral  equations, 
the  number  of  roots  of  a  given  irrational  equation. 

9.  Often  no  solution  of  the  derived  rational  equation  will  satisfy 
the  given  irrational  equation,  if  the  roots  are  restricted  to  principal 
values  only. 

Ex.  4.    Solve                         I  -  ^/x+  ^2x-\-l  =  0.  (1) 

Transposing,                                          1  +  '\/2x  +  1  =  -y^.  (2) 

Squaring  both  members,  1  +  2\/'2x  +  1  +  2ic  +  1  =  a?.  (3) 

Collecting  terms  and  transposing,             2^2  x  +  1  =  —  2  —  x.  (4) 

Again  squaring  both  members,                 4  (2a:  +  1)  =  4  +  4a;  +  a;^.  (5) 

The  equivalent  equation                                x^  —  4  a:  =  0,  (6) 
has  as  solutions                x  =  0,             and               a:  =  4. 

We  find  by  substitution  that  neither  of  these  values  satisfies  the  given 
equation  (1).     Hence  equation  (1)  has  no  root. 

It  may  be  seen  that,  if  the  signs  of  the  terms  of  the  given  equation  are 
changed  among  themselves  in  all  possible  ways,  we  shall  obtain  the  set  of 
four  irrational  equations  which  all  lead,  when  rationalized,  to  the  rational 
equation  (6).     (See  §  7,  (1),  (2),  (3),  (4),  (5).  ) 

Substituting  the  values  a;  =  0  and  a:  =  4  in  these  equations,  we  find  that 
(i.)    Neither  value  satisfies  the  given  e([uation 

1  -  V^  +  /v/2  a3  +  1  =  0,  (1) 

(ii.)    One  value  x  =  0,  satisfies  the  additional  equation 

1  _  y^  _  .y/2a:-|-  1  =  0,  (7) 

(iii.)    Neither  value  satisfies  the  additional  equation 

I  +  \/x  +  \/2x-\-  I  =  0,  (8) 

(iv.)    Both  values  satisfy  the  additional  equation 

-  I  -  V^  +  V^x  +  l  =  0.  (9) 

10.  Certain  irrational  equations  may  be  so  written  as  to  appear  in 
quadratic  form  with  respect  to  some  expression  appearing  in  them, 
and  the  principles  for  the  solution  of  quadratic  equations  may  then 
be  applied  to  obtain  their  solutions,  if  indeed  any  solutions  exist. 

Ex.  5.    Solve  x2  +  53:-  2^./.=^  +  5 ^Ts  -12  =  0.  (1) 


532  FIRST  COURSE   IN  ALGEBRA 

Observe  that  we  may  duplicate  the  expression  x^  +  5  .r  +  3,  which  appears 
iiiuler  the  radical  sign,  by  adding  3  to  the  expression  x'^  +  bx  which  appears 
outside. 

Accordingly,  adding  3  and  also  subtracting  3  from  the  first  member  of  (1), 
we  obtain  the  equivalent  equation 

a;2  +  5 a:  +  3  -  2^/x^ -\- bx  +  S  -15  =  0.  (2) 


Factoring  with  reference  to  >y/a:^  +  5  a:  -f  3,  we  obtain 

[V'a:--»  +  5a:+3  -  bj_.y/x^  +  5  x  +  3  +  3]  =  0.  (3) 

The  roots  of  equation  (3)  include  all  of  the  roots  of  the  ftdlowing  equations: 

/y/x^  +  5  X  -I-  3  -5  =  0,    (4)      and     ^/x^T~5x~^  +  3  =  0.  (5) 

The  roots  of  e^iuation  (4)  are  found  to  be  x  =  ^^^ .     These 

values  will  be  found  by  substitution  to  satisfy  the  given  equation. 
From  equation  (5)  we  obtain 

y^x2  +  5x  +  3  =  -  3.  (6) 

Hence,  x^  _^  5  a; -|_  3  =      9.  (7) 

Or,  x«  +  5  X  -  6  =      0. 

Factoring,  (x  +  6)(x  —  1)  =      0. 

Hence,  x  =  —  6,  and  x  =  1. 

Since  the  right  member,  9,  of  equation  (7)  was  obtained  by  squaring  a 
negative  number  in  the  second  member  of  equation  (6^,  it  follows  that  when 
the  values  —  6  and  +  1  are  substituted  for  x  in  the  expression  x^  4-  5  x  +  3 
appearing  under  the  radical  sign  in  equation  (6),  the  expression  will  repre- 
sent the  square  of  a  negative  number.  Accordingly,  when  finding  the  value 
of  y\/x^  +  5x  +  3,  after  having  substituted  —  6  and  +  1  for  x,  it  is  necessary 
to  consider  that  the  result  is  a  negative  number. 

With  this  understanding,  it  may  Ixi  seen  that  the  values  —  6  and  +  1 
satisfy  the  given  equation. 

For,  substituting  —  6,  we  have,  36  —  30  —  2(—  3)  —  12  =  0.     Hence,  0  =  0. 

Substituting  -  1,  we  have,  1  +  5  -  2(-  3)  -  12  =  0.     Hence,  0  =  0. 


Ex.6.    Solve2x2-2x- /^x2-x  +  4  4-2  =  0.  (1) 

Equation  (1)  may  be  written  in  the  equivalent  form 


2x2-2x  +  8- >y/x'«-x  +  44-2-8  =  0.  (2) 

Or,  2(x2  _  X  +  4)  -  yy/^^^T^T^  -6  =  0.  (3) 

Equation  (.3)  is  quadratic  with  respect  tx)  the  expression  /y/x^  —  x  +  4. 
Hence,  factoring  with  reference  to  -y/x^  —  x  +  4,  we  have, 

(2>v/x2  -  X  +  4  +  3)(y'x2-x  +  4-  2)  =  0.  (4) 


IRRATIONAL   EQUATIONS  533 

Placing  these  factors  equal  to  zero,  we  obtain  the  equations 

2/y/x''«  -  a;  +  4  +  3  =  0,  (5)  and  ^/x^  -  a:  +  4  -2  =  0.  (6) 


From  equation  (5)  we  obtain  /\/x2^^^^^  +  4  =  —  f ,  the  solutions  of  which 

are  found  to  be  x  = ^- 

It   shouhl    Ije   observed   that  when  these  values  are  substituted   for  x. 


'y/x'^  —  X  -\-  4  must  be  a  negative  number,  —  f . 

Accordingly,  with  this  understanding,  these  imaginary  values  will  be 
found  to  satisfy  the  given  equation. 

Equation  (6)  is  found  to  have  the  solutions  a;  =  0  and  a:  =  1,  both  of 
which  satisfy  the  given  equation. 

Exercise  XXIV.    1 

Solve  the  following  irrational  equations,  verifying  integral  or 
fractional  results  and  rejecting  "  extra  roots  " : 


1.  V^  +  a;"'  —  3  =  0.  6.  VaJ  +  6  =  ^'i\x  —  H  —  2. 

2.  Vx^  +  15  —  7=0.  7.  a/25 -a?  +  Vl6.  +  x  =  9. 

3.  'v/3ic+  4  +  V2x-\-  9  =  1.      8.   Vx -\-  13  +  V2ic  — 45  =  7. 

4.  Vie  +  X  +  Vl  -«  =  5.         9.   Vx-\-  5  +  V2a;  +  8  =  7. 

5.  VSx+lo  =  VSiB-  1  -  1.   10.  V''2x+  1  +  Va;+  5  =  6. 

11.  «\/ic^  +  20  H-  a;\/a;2  +  10  =  5. 

12.  (a)  2  +  ^30;  — 2  —  V8^  =  0. 

(b)  2  —  VSx-2  —  VSx  =  0. 

(c)  2  —  V3  a;  —  2  +  V^  =  0. 

(d)  V3a;—  2  —  2  -  VSaj  =  0. 

13.  (a)    Vx+  2  +  Va;—  13  +  Va;  —  5  =  0. 

(b)  Vi«4-  2  -  Va;—  13  —  ^/x-  6  =  0. 

(c)  a/«+  2  -  Va;  -  13  +  V^c  -  5  =  0. 

(d)  ViB+  2  +  Vaj-  13  —  Vx  —  5  =  0. 

14.  Va;+  12  +  Va;—  12  -  Vx+  23  =  0. 

15.  V5  —  a;  +  V2  +  a;  =  Vl4. 

16.  a/5  — a;  +  \/8  -  a?  =  Vl3  —  2  a;. 

17.  ^2  a;  +  9  —  Vaj  —  4  =  Va^  +  1. 

18.  ^5  +  2a;  =  a/3  +  a;  +  a/2  +  a;. 


534  FIRST  COURSE  IN  ALGEBRA 

19.  V4-a;  =  V^  +  x  +  V9  +  a. 

20.  V2aj+  9  +  Va  — 4  =  Va;  +  41. 

21.  V'^x+  1  +  Va—  1  =  V2 a;  —  6. 

22.  VSx  +  4:  —  \/2x+  6  =  Va;  -  14. 

23.  V(a;  ~  4)(x  ~~3)  +  V(^  -  2)(x  -  1)  =  V2. 

24.  V(4  +  «)(«  +  1)  +  V(4:  -  x)(x  -  1)  =  4v^. 

25.  Va  +  «  —  Va  —  x=  Va. 

26.  Va=*  +  X  +  Vb^-x  =  a  +  b. 

27.  Va  —  x  +  Vft  — aj  =  a/«  +  ^  —  2 

28.  '^S^^^^  - 'iJ'a'^^  =  V^6'=r^. 
3 


Ia% 


29. 


Vx  —  4 


+  V«  -  4  =  4. 


30.   ^^+^^"-^  =  1. 
3  a;  —  V3  a;  —  3 


31 


1 


X  +  Va^—  1      ic  —  Var^  —  1 
32 
33.   2A/2a  +  \/2a;+  9  = 


V2a;  — 3       VSa;  — 2_       7 
A/3a;-  2       V2a;--3~       12 

65 


^20;+  9 


34.  Vx  +  2+     ^  =  a;  +  3. 

Vx  +  2 

_.    x+  1      6+  1 

35.  — -=-  =  — —  • 

yx        yb 

a  —  X  x  —  b 


36.  -:^^  +  ^=^=v^^r6. 

V  a  —  X      yx  —  6 
37  I  _  I  _k 

^  _  VF  -  ar^      ^  +  v^F=^^      a^ 

38.    i/^_v/^  =  v/^_v/^. 
Vic       Va;       Vw       Vw2 


IRRATIONAL  EQUATIONS  635 


Va^  +  x^  +  Va^  —  x^  _  Va  +  Vc 

OJ.     ■  .  —        "z: IT' 

V  «^  +  x^  —  wd^  —  Q?       V  a  —  Vc 

40.  Va;'  -  a;  +  1  -  V^M^^T+T  =  c. 

41.  aj2  —  3  a;  —  Va?"'  —  3  £«  —  2  =  0. 

42.  VlO  -ar'  —  a;  =  «  —  a;2  —  a. 


43.    2a;2  +  a;  +  A/2a;'  +  a;  -  42  =  0. 


44.  2  cc^  +  6  aj  +  Va;'  +  3  a;  =  10. 

45.  2 a;'  +  a;  -  3V2ar^  +  aj+  4  =  6. 


46.  2a;'-^  —  10a;  +  12  —  2Va;'  —  5a;  +  8  =  0. 

47.  3ar*  — 4aj+  V3a;'-4a;  — 6  =  18. 

11.  At  least  one  solution  of  certain  equations  which  have  the 

p 
special  form  x''  =  a  may  be  obtained  by  the  following  process  : 

p 
From  the  equation  a;*"  =  a, 

p   r  r 

we  obtain,  {x'Y  =  dP. 

r 

Therefore  x  =  aJ^. 

12.  It  should  be  observed  that  more  solutions  exist  than  are 
commonly  obtained  by  the  process  above. 

Ex.  1 .    Find  one  or  more  solutions  of  x^  =  3. 

From  x^  =  2, 

we  have,  (x^y  =  3^. 

That  is,  x  =  9. 

In  this  case  the  solution  obtained  is  the  only  one  which  exists  tor  the 
given  equation. 

Ex.  2.    Find  one  or  more  solutions  of  y^  =  8. 
From  y^  =  8, 

we  have  (y^r  =  ^  • 

That  is,  y  =  ^^64  =  4. 

The  solution  obtained,   y  =  4,   is  in  this  case  one  of  the  three  possible 
solutions  of  the  given  equation. 


536  FIRST  COURSE   IN  ALGEBRA 

The  complete  solution  of  the  equation  may  be  obtained  as  follows : 

From  y^  =  8, 

we  obtain  2/*  =  ^4. 

Hence  y«  _  64  =  0. 

Factoring  (y  -  4)  (y^  .f  4  ?/  +  1  '0  =  0. 

Solving  the  equations  obtained  by  i>ljicing  these  factors  separately  equal 
to  zero,  we  have 
from  y  —  4  =  0,         and  from  y^  +  4y  -{■  IG  =  0, 

2/  =  4,         and  2/  =  -  2  db  S/y/^. 

Accordingly,  the  three  solutions  of  the  given  equation  are  the  real  nmnher 
4  and  the  conjugate  complex  numbers  —  2  +  2^'—  3  and  —  2  —  '2,^/—  3. 

These  solutions  will  be  found,  by  substitution,  to  satisfy  the  given 
equation. 

Mental  Exercise  XXIV.    2 

Obtain  one  or  more  solutions  of  each  of  the  following  equations, 
regarding  a,  y,  z,  and  w  as  unknowns  and  all  other  letters  as  repre- 
senting known  numbers  : 


1.  x^  =  2. 

17. 

rz=h 

33. 

x^  =  64. 

2.  a*  =  3. 

18. 

x^  =  b\ 

34. 

x^  =  32. 

3.  a^  =  2. 

19. 

x^  =  c*. 

35. 

.^  =  -1. 

4.  jc*  =  -  3. 

20. 

y^  =  ab\ 

36. 

x^  =  S. 

5.  y^  =  -  6. 

21. 

y^  =  kV. 

37. 

x^  =  a\ 

6.  -.^^  =  4. 

22. 

z^  =  ab\ 

38. 

x^  =  a\ 

7.  -y^  =  6. 

23. 

qi^  =  —  bc^. 

39. 

S  =  a\ 

8.  V^  =  5. 

24. 

W^  =  4:. 

40. 

y^  =  b\ 

9.  v^=7. 

25. 

x^  =  27. 

41. 

z^  =  a-\ 

10.   ^z  =  4.. 

26. 

xi  =  8. 

42. 

y^  =  nr\ 

11.  ^w  =  S. 

27. 

/  =  27. 

rt^ 

12.  v^  =  -4. 

28. 

yi  =  S. 

43. 

J=±. 

13.  v^  =  3. 

29. 

z^  =  16. 

14.  ^x  =  2. 

30. 

z^  =  9. 

44. 

4      a'    • 

15.   </x=l. 

31. 

^^=36. 

9 

16.   \/y  =  -2. 

32. 

^/=16. 

45. 

V2^  =  4. 

IRRATIONAL  EQUATIONS 


537 


46. 
47. 
48. 
49. 
50. 
51. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 

59. 
60. 

61. 
62. 
63. 
64. 
65. 
66. 
67. 
68. 
69. 
70. 
71. 
72. 
73. 


V^y  =  6. . 
^57=  10. 
a/22^  =  8. 
^Ix  =^  -  2. 
\)^2a'  =  2. 
A^4^  =  4. 
a/6^  =  2. 
VHjrc  =  5. 
a/3  w  =  4. 
\/5w  =  6. 
x^  =  h 

zi  =  h 

J  =  '-^ 
a 

1       a 

V'W  =  ^. 
Vx  =  h 
\/x  =  i 
V^  =  — i. 
2a;^  =  3. 


3aj^  =  4. 
2ic*  =  5. 
3  a;*  =  2. 
2a;^  =  5. 

3V^=2. 

3V^  =  4. 


74.  4a/^  =  1. 

75.  5V^=6. 

76.  2a/3^=5. 

77.  3V2^  =  2. 

78.  4a/5w  =  5. 

79.  V«ic  =  ^. 

80.  Vcx  =  d. 

81.  V%"=  1. 

82.  a\/w  =  b. 

83.  ca/^  =  d. 

84.  c?V^  =  1. 

85.  aVx  =  2. 

86.  V^  =  7^. 
T  m 

87.  \/^  =  6. 
▼  a 

88.  V~z  =  ^- 


89.  v^ 


90.^=1. 
a 

91.  a/^=2. 

92.  v':?/  =  3. 

93.  -5^21  =  1. 

94.  a/2^=:3. 

95.  a/3S  =  3. 

96.  "-^3;=  2. 

97.  x^  =a-\-b. 

98.  2/^  =za  —  b. 

99.  ;j^  =  c  +  ^. 


w^  =  m  —  n. 


W: 


02. 


03. 


04. 


05. 


06. 


07. 


2 
\/x 


v^ 


=  3. 


=  3. 


=  5. 


3    ~^- 


2V/-  =  7. 


08.  - 
09. 

10. 

11. 
12. 
13. 
14. 

15. 


3V  2 

2 
3V  5 

a/3^ 
3 

5 

i_ 


=  1. 


=  2. 


^  ^  =  6. 
z~^  =  3. 

;5"^  =  2. 

Jl  — 1. 

ii?*      2 

1-1. 

x^      3 


16.  a;  ^  = 


17.  ^J 


-i 


538  FIRST  COURSE   IN   ALGEBRA 

llg    ;j-i  =  1.  121.  z^  =  2K  124.   \^x  =  Ve. 
^                               —           —  _ 

^,^      1         1  122.   Vy='V^5.  125.   V3^  =  2'^5. 

119.  a^^  =  o*. 

120.  V«  =  Vq.  123.  «<;*  =  5^  126.  Vz  =  3v^7. 

Special  Equations 

13.  We  will  now  consider  certain  special  equations,  the  solutions 
of  which  depend  upon  the  solutions  of  quadratic  or  of  linear  equations. 

Certain  ecjuations  containing  two  different  powers,  tc^"  and  a;",  of 
an  unknown  quantity  or  expression  x,  one  power  being  the  square  of 
the  other,  may  be  reduced  to  the  form  ax^"  +  hx^  +  c  =  0  which  is 
quadratic  with  reference  to  af*.  Such  equations  may  be  solved  by 
the  methods  employed  for  the  solution  of  the  standard  quadratic 
equation. 

E.g.  The  following  equations  are  all  in  quadratic  form,  since  in  each  case 
the  power  of  the  unknown  appearing  in  the  first  term  is  the  square  of  the 
power  appearing  in  the  second  terra  : 

a:4  _  25  x2  +  144  =  0, 

a; -14x^  +  45  =  0, 

a;^- 13x^  +  36  =  0, 

ar2  -  ari  -  6  =  0. 

Ex.  1.  Solve  x*  -  25  a;2  4-  144  =  0.  (1) 

Factoring,  (x^  -  16)(x2  -  9)  =  0.  (2) 

This  equation  is  equivalent  to  the  set  of  two  quadratic  equations 

ar2_i6  =  0     and     x^  -  9  =  0.  (3) 

Hence  x  =  ±  4     and  x  =  ±  3. 

All  of  these  values  will  be  found   by  substitution  to  satisfy  the  given 
equation. 

Ex.  2.  Solve  X  -  14  xi  +  45  =  0.  (1) 

Observe  that  x,  which  appears  in  the  first  term,  is  the  square  of  x^  which 
appears  in  the  second  term. 

Factoring  with  reference  to  x^,  we  obtain, 

(x^  -  9)(x2  -  5)  ^  0.  (2) 


SPECIAL  EQUATIONS  539 

This  single  equation  is  equivalent  to  the  set  of  equations 

a;^  —  9  =  0,       and     x^  _  5  =  0.  (3) 

Hence,  x^  =  9,       and  x^  =  5. 

Squaring  both  members  of  each  of  the  equations  above, 

(x^y  =  92,      and       (x^y^  =  52. 
Therefore  a:  =:  81,     and  x  =  25. 

These  values  will  be  found,  by  substitution,  to  satisfy  the  given  equation. 
Ex.  3.    Solve  a;^  -  I3x^  +  36  =  0.  (1) 

Factoring  with  reference  to  a:^,  we  have 

(a:f  _9)(a;t_4)  =  0.  (2) 

Hence,  a;^  -  9  =  0,  and  art  _  4  =  0.  (3) 

Tlierefore,  x^  =  9,  and  x^  =  4. 

To  obtain  x  from  a:»,  we  may  raise  x^  to  the  third  power  and  extract  the 
square  root  of  the  result,  or  first  extract  the  square  root  of  x^,  and  then  find 
the  third  power  of  the  result. 

That  is,  (xt)t=:9^  and  Gf)t  =  4^. 

Therefore  a;  =  i  27,  and  x  =  ±H. 

These  values  will  be  found,  by  substitution,  to  satisfy  the  given  equation. 

Ex.  4.    Solve  ar2  -  a;-i  -  6  =  0.  (1) 

Factoring  with  reference  to  xr\ 
we  have  (xr^  -  3)(a;-i  +  2)  =  0.  (2) 

Hence,  x-^  -  3  =  0,  and        a;-i  +  2  =  0.  (3) 

We  find  that  ar^  =  3,  and  a:-i  =  -  2.  (4) 

To  obtain  x  from  xr\   we  may  write   (x~^)~^  =  x+^. 
Hence,  (a:-i)-i  =  3-1,        and         (a;-i)-i  =  -  2-\ 

Therefore,  ^  =  h  ^^^^  ^  =  —  \- 

Instead  of  proceeding  as  above,  we  may  obtain  the  values  as  follows : 
From  ari  =  3  and  x-^  =  -  2, 

we  obtain  -  =  3  and  -  =  —  2.  (5) 

X  X 

Hence,  a:  =  ^  and  x  =  —  \. 

Verifying  by  substituting  in  (1),  we  have, 


540  FIRST  COURSE  IN   ALGEBRA 


Substituting  |^, 

Substituting  —  i, 

a)-^-a)-'-6=o 

(-h)-'-(-h)-'-(i  =  o 

9     _     3     -6=0 

4+2        -6=0 

0  =  0. 

0  =  0. 

14.  If  the  solution  of  a  given  equation  cannot  be  readily  obtained 
by  factoring,  we  may  either  resort  to  the  method  of  completing  the 
square,  or  we  may  use  the  formula. 

Ex.  5.   Solve  3x-Hxi  +  2  =  0. 

Observe  that  x  in  the  first  term  is  the  square  of  x^  in  the  second  term. 

Referring  to  the  standard  equation  ax^  +  ftx  +  c  =  0,  we  find  that  x  and 
a:*  in  the  given  equation  are  represented  by  x^  and  x  respectively  in  the 
standard  equation;  the  coefficient  3  of  a:  in  the  given  equation  is  repre- 
sented by  the  coefficient  a  of  x^  in  the  standard  equation  ;  —  8  is  repre- 
sented by  6,  and  2  is  represented  by  c. 

Corresponding  to  the  solution  of  the  standard  equation, 


_-h  ±  ^b^-4ac 
^-  2a 


I      -  (-  8)  ±  V(- 8)^^-4.  3"^ 
we  may  wnte  x'  =  — ^^ 5^-^— 


_  +  8  j-  V64  -  24 
~  6 

+  8  ±  2\/l0 
~  6 

_  +  4  +  yio 

~  3  * 

,       .       "          i      +  4  +  VlO              ,          1      +  4  -  a/To 
That  is,  a:^  =  ^    —         and       x^  = ^ 

Squaring  both  members  of  each  of  the  equations  above, 

26  +  8a/10              ,                  26-8/v/lO        *    . 
Or,  x  =  -—^-^ — ,       and         x= ^ 

These  exact  values  will  be  found  by  substitution  to  satisfy  the  given 
equation. 

By  extracting  the  square  root  of  10  to  any  required  number  of  significant 
fifnires,  and  replacing  /y/lO  by  the  approximate  value  thus  found,  approxi- 


SPECIAL  EQUATIONS  541 

mate  values  may  be  obtained  for  x,  which  are  correct  to   any  required 
number  of  significant  figures. 

Thus,  X  =  5.6997+,         and         x  -  .0780+.     (See  Chap.  XXII. 

§  20,  Ex.  2.) 

Exercise  XXIV.    3 

Find  one  or  more  solutions  of  each  of  the  following  equations, 
verifying  all  integral  and  fractional  solutions:    ; 

1.  x'  -  U)x'  +  9  =  0.  17.  4^1  _  i^J  +5  =  0. 

2.  ;«^- 200^^+64  =  0.  13.   14a^^  =  1107-a^. 


3.  x'  -14x^  =  —  1225. 


19.  Sx^  -  4x^  =  160. 


4.  0.^3  0.^ +19)  =124.  20.  Sx^-l  =  4xK 

5.x^-(x  +  2r  =  0.  21..U.^  =  756. 

6..^=(.+  6y.  22.  5-3.-  =  2.-. 

7.  16(a;^-3)=a.^  23    x'^  -  x'^  =  Q 

8.  (,+  12)^_,.  =  0.  24.  4.--32  +  .-  =  0. 

9.  2.«  +  48  =  .«.  25.  20.-^-.-^  =  64. 


10.  ««  +  9ic»  +  8  =  0. 

11.  5a;^  — 2a;  =  — 3. 


26.  x^-5x''=  Ux. 

Hint.  Write  in  the  form 


12.  x+  5x^  =  36.  x(x^  -  5x  -  14)  =  0, 

13.  3  £C^  —  5  a  =  —  36.  Then  a;  =  0  is  one  solu- 

14.  x^  +  4a;^  =  21.  ^io"- 

15.  x^  (3  x^  —  2)  =  8.  27.  x(x'  -  16)  =  45(£c  —  4). 

16.  2a;^  -  2  =  SxK  28.  a;^  —  Gx^  -  40aj^  =  0. 

15.  Occasionally  the  terms  of  an  equation  of  degree  higher  than 
the  second  may  be  so  grouped  as  to  allow  of  the  reduction  of  the 
equation  to  a  form  which  is  quadratic  with  respect  to  some  definite 
group  of  terms  containing  the  unknown. 

If  we  let/(x)  represent  a  group  of  terms  containing  the  unknown, 
X,  we  may  represent  an  equation  which  is  quadratic  with  reference 
to  this  group  of  terms  by 

a[/(x)r  +  b[f(x)]  +  c  =  0. 
Ex.  1.    Solve  (a;2  +  3  x)2  -  2(a;2  +  3  a;)  _  8  =  0.  (1) 

Observe  that  the  equation  is  quadratic  with  respect  to  the  group  of  terms 
(x2  +  3a:). 


642 


FIRST  COURSE  IN  ALGEBRA 


(-4.0) 


Factoring  with  respect  to  this  group  of  terms,  we  obtain, 

[(x2  +  3  a:)  -  4][a;2  +  3x)+2]=0.  (2) 

Equation  (2)  is  equivalent  to  the 
set  of  two  equations  obtained  by- 
writing  the  factors  of  its  first  member 
separately  equal  to  zero. 

Hence,     a:^  +  3  a;  -  4  =  0,         (3) 

and  a:2  +  3a;  +  2  =  0.        (4) 

These  equations  in  turn  are  equiv- 
alent to 

(x  +  4)(x-l)  =  {), 

and  (x  +  2)(x  +  I)  z=  0. 

Hence,       ar  =  —  4,  a:  =  1, 


(6) 
(6) 


and 


ar  =  -2,  a;  =  -l. 


Fig.  1.    (T2  +  3x)2-2(a:2-f  3a:)-8  =  .y. 


These  values  are  all  found  by  sub- 
stitution to  be  roots  of  the  given 
equation,  which  is  of  the  fourth  de- 
gree or  biquadratic.     (See  Fig.  1.) 

Ex.  2.   Solve 


a:*-12a:8  +  25a;2+66a:-80  =  0.    (1) 

We  may  obtain  the  tenn  which  is  necessary  to  complete  the  square  with 
respect  to  x*  —  12  a:*  by  using  —  12  jc^  as  a  "  finder  term  "  with  a;*,  as  follows  : 

-12a:8 


2a;2 


=  -6x. 


Accordingly  the  complete  trinomial  square  of  which  x^  and  —  12  a:*  are 
the  first  two  terms  is 


x*-l2x»-\-  (-6xy 


12a;8  +  36a;2. 


This  may  be  constructed  in  the  first  member  of  (1)  by  adding  11  a;^  to 
25a:2. 


Hence, 

Or, 

Hence, 


c*-  12a:«  +  25x2+  11  a;^-  11  cc^  +  66x  -  80  =  0. 

X*  -  12a:8  +  36  3.2  _  n  a:^  +  66a:  ~  80  =  0. 

{a:2  -  6  a:)2  _  11  a:2  +  66  a:  -  80  =  0. 


(2) 
(3) 
(4) 


To  arrange  the  terms  in  quadratic  form  with  respect  to  a;^  —  6  x,  we  must 
so  group  them  as  to  have  both  (a:^  —  Gx)'^  and  (x^  —  6  a:). 


SPECIAL  EQUATIONS  543 

Grouping  the  terms  —  11  a:^  and  66  a;,  we  have 

(a;2_6;r)2- 11  (a:2-6a:)-80=:0.  (5) 

We  have   derived  from   equation   (1)  the  equivalent  equation  (5)  in 
quadratic  form. 

Solving  (5)  by  the  method  of  factoring,  we  have 

[(a:2  -(5x)-  16][(a:2  _  6 a;)  +  5]  =  0.  (6) 

This  equation  is  equivalent  to  the  set  of  two  equations 

a;2  _  6  a;  -  16  =  0,    (7)      and      a:2  _  6  a:  +  5  z=  0.     (8) 
The  solutions  of  these  equations  are  found  to  be 

a:  =  8,  a:  =:  —  2,  and  a:  =  5,  a:  =  1 . 

All  of  the  values  are  found  by  substitution  to  be  solutions  of  the  given 
equation  of  the  fourth  degree. 

16.    The  introduction  of  an  auxiliary  letter  in  place  of  a 

term  or  group   of  terms  containing  the  unknown  often  simplifies 

the  process  of  solution  of  an  equation. 

a;2  +  6         5  a;         „ 
Ex.3.    Solve  ___  +  _^__  =  6.  (1) 

Observe  that  in    the  given    equation    the  expression    and  its 

X 

reciprocal     ^  both  appear. 

a:^  +  6  X  1 

If  we  let =  y,  then  its  reciprocal  may  be  written  -^ — -  =  -  • 

X  X  "^  o      y 

For  (1)  we  may  substitute,  ^  +  -  =  6,  (2) 

from  which  y^  —  6y  +  5  =  0.  (3) 

We  find  that  2/  =  5,     (4)     and  y=l.     (5) 

Substituting  these  values  for  y  in =  y  we  shall  obtain  the  two 

following  equations,  which  are  together  equivalent  to  equation  (1)  : 
?1±^  =  M6)  and  ^-if5=l.(7) 

The  solutions  of  these  equations  are  found  to  be 


1  ±  V-  23 
a:  =  3,  a;  =  2,  and  x  = 1 

These  values  all  satisfy  the  given  equation. 

Ex.4.    Solve  (a;2  4-a:  +  2)(a:2  +  a:  +  7)  =  36.  (1) 

If  we  let  a:2  +  a:  +  2  =  y, 

tlien  x^  +  x  +  7  =  y-\-5. 


544  FIRST  COURSE  IN  ALGEBRA 

Substituting  y  and  y  -\-  5  for  the  polynomials  in  the  given  equation, 
we  have  y  (y  +  5)  =  36.  (2) 

Or,  2/2 +  52/ -36  =  0.  (3) 

The  solutions  of  this  equation  are  found  to  be 

!/  =  -9,  (4)  and  y  =  4.  (5) 

Substituting  these  values  for  y  in  the  assumed  equation  x^  +  x  -{-  2  =  y^ 
we  obtain  the  two  following  equations,  which  are  together  equivalent  to 
equation  (1)  : 

a;2  +  a:  +  2  =  -  9,         (6)         and         x^ -\- x  +  2  =  4.         (7) 
The  roots  of  these  equations  are  found  to  be 

X  = ~ ,  and  a;  =  —  2,  a:  =  +  1. 

These  values  will  be  found,  by  substitution,  to  satisfy  the  given  equation. 
Examples  3  and  4  may  also  be  solved  by  the  method  employed  for 
examples  1  and  2. 

Exercise  XXIV.    4 

Solve  the  foUowiDg  equations,  verifying  all  integral  and  fractional 

solutions  : 

1.  (x"  -Sxy-  six"  -^x)  =  20. 

2.  (x"  +  xy  -  2G(x^  +  x)  +  120  =  0. 

3.  3ar'+2ic+  1  =77-3 


4.  £c2  -  4a;  -  26  +  ^ —  =  0. 


Sx^  +  2x 

105 
x^  —  4x 

5.  X*  +  2x'^-Gx^-~  7£c- 60  =  0. 

6.  x*-2x^  +  Qx''-5x=  14. 

7.  X*  +  (Jx^  +  Ux^  +  15ic  +  6  =  0. 

8.  X*  —  10a;*  +  14a;2  +  55a;  +  30  =  0. 

9.  (x^-x-  4)(x^  -  aj  -  3)  -  6  =  0. 
10.  (ar^  +  a;  4-  l)(a;'  +  a;  +  3)  =  63. 

11. 
12. 


X 

1 

x^- 

X 

35  _ 
6  ~ 

1  _ 

X 

73 

x" 

7? 

-  1 

+  2 

1 
1 

24 
-3 

X 

-3 

1 

+  2 

SPECIAL  EQUATIONS 


645 


-  i'-iy 


+  -  +  a;  =  42. 

X 


x^+  1 


=  12. 


+  Sx 


24 


=  10. 


17.  A  binomial  equation  is  an  equation  of  the  form  cc"  +  a  =  0, 
in  which  ?i  is  a  positive  integer.  Whenever  the  binomial  x""  +  a 
can  be  expressed  as  the  product  of  two  or  more  factors  of  the  first 
or  second  degrees,  the  binomial  equation  x""  -\~  a  =  0  may  be  solved 
by  the  methods  employed  for  the  solution  of  linear  and  quadratic 
equations. 

Ex.  1.    Find  the  three  cube  roots  of  +  1. 

If  X  represents  any  one  of  the  cube  roots  of  +  1, 
we  have 

a:8  =  +  l.  (1) 

By  solving  the  binomial  equation,  we  shall  find 
the  three  cube  roots  desired. 

Equation  (1)  is  equivalent  to  a;*  —  1  =  0.       (2) 

Factoring,  (x  -  l)(x2  +  a:  +  1)  =  0.      (3) 

This  equation  is  equivalent  to  the  set  of  two 
equations 

a;  _  1  -  0,  (4)     and     x2  +  x  +  1  =  0.  (5) 

The  solutions  of  (4)  and  (5)  are 

-1±  v^ 


K{-i+V=3) 


K(-i-V=3 


Fig.  2.    x^  -  1  =  y. 


x=l 


and 


(See  Fig.  2.) 


The  positive  real  value  x  =  +  1  is  the  principal  root  of  +  1.  This  is  the 
root  found  by  the  arithmetic  process  for  the  extraction  of  the  cube  root. 

The  three  values  obtained  above  will  all  be  found  to  satisfy  the  algebraic 
equation  (1). 

18.  An  equation  in  which  the  coefficients  are  the  same,  whether 
read  in  order  forward  or  backward,  is  called  a  reciprocal  equation. 

Ex.  2.    Solve  the  binomial  equation  x^  —  1  =  0.  (1) 

Factoring,  (x  -  l)(r*  +  ar*  +  x^  +  x  +  1)  =  0.  (2) 

This  single  equation  is  equivalent  to  the  binomial  equation 

x-l=  0,  (3) 
35 


546  FIRST  COURSE   IN  ALGEBRA 

and  the  reciprocal  equation  x*  -{■  x^  +  x^  +  x  +  I  =  0,  (4),  taken  together. 
The  solution  of  (3)  h  x  =  1. 

The  solution  of  the  reciprocal  equation  (4)  may  be  obtained  as  follows  : 
Dividing  each  of  the  terms  of  (4)  by  x%  we  obtain 

a:2  +  X  +  1  +  ^  +  i^  =  0,  (5) 

or,  •  a:2  +  1+  L  +  a:  +  i  =  0.        •  (6) 

X  X 

The  terms  x^  and  1  /x^  suggest  a  trinomial  square,  a;^  _|_  2  -^  __. 

By  adding  1  and  also  subtracting  1  in  the  first  member  of  equation  (6), 
we  obtain, 

ar«  +  2  +  i  +  a:  +  i-l  =  0,  (7) 


or, 


Equation  (8)  is  quadratic  with  respect  to     (^  +  ~  )  * 
Accordingly,  applying  the  formula,  we  obtain 

X  z 

Equation  (9)  is  equivalent  to  the  set  of  two  separate  equations 

From  these  we  derive 

:.._(-l  +  V^')x+1^0    and    x=- (^i^)x  +  1  =  0. 

These  four  complex  values,  taken  together  with  the  real  value,  a;  =  1, 
obtained  from  equation  (3),  are  the  five  fifth-roots  of  +  1,  which  are  the 
solutions  of  the  given  equation  a^  —  \  =0. 

Exercise  XXIV.     5 

Solve  the  following  equations,  verifying  all  integral  and  fractional 
solutions  : 

1.  ic«  -  27  =  0.  3.   x^  =  1. 

2.  aj^  -  64  =  0.  4.   a?«  +  1  =  0. 


PROBLEMS   IN   PHYSICS  547 

5.  ic^  +  8  =  0.  7.    (x  -  sy  =  8. 

6.  a;«  -  64  =  0.  8.    (x  -  4/  -81  =  0. 

9.  6a?* -35a^  + 62a;2  — 35£C+ 6  =  0. 

10.  lOx*— 11  x^  +  IbOx^  —  nx-^-  10  =  0. 

11.  6a;*-49a;*  + 86a;'  — 49a;+ 6  =  0. 

12.  6a;*  — 25ic^4- 38a;'- 25£c+ 6  =  0. 

13.  12a;*  -  91  a;«+  194a;'- 91a;  +  12  =  0. 

14.  15a;»  -  49a;' +  49a;- 15  =  0. 

15.  8  a;^  —  46  a;*  +  47  a;'  +  47  a;'  -  46  a;  +  8  =  0. 

Problems  m  Physics 

19.  The  Simple  Pendulum.  If  a  simple  pendulum  swings 
through  an  arc  the  extremities  of  which  are  A  and  B,  the  motion 
of  the  pendulum  from  A  to  B  or  from  B  to  A  is  called  an  oscillation 
or  a  vibration. 

20.  The  time  of  vibration  is  the  time  required  for  the  pendulum 
to  make  a  single  vibration,  that  is,  to  move  from  A  to  B^  or  from 
B  to  A. 

21.  The  distance  from  the  lowest  point  of  the  arc  AB  to  either 
extremity,  A  or  B,  that  is,  one-half  of  the  arc  AB,  is  called  the 
amplitude  of  vibration. 

22.  When  the  amplitude  of  vibration  is  very  small,  an  approxi- 
mate value  of  the  time  of  vibration,  t,  expressed  in  seconds,  of  a 
pendulum  of  length  /,  expressed  in  feet,  is  found  by  the  formula 


^  g 


In  the  following  examples  the  approximate  value  22/7  maybe 
taken  for  the  numerical  constant  tt. 

Exercise  XXIV.     6 

Solve  each  of  the  following  problems  relating  to  the  simple 
pendulum  : 

1.  Find  the  length  in  feet  of  a  pendulum  which  vibrates  once  in  a  second 
at  a  place  at  which  g  =  32.16. 

2.  Find  the  value  of  g  at  a  certain  place  if  a  pendulum  which  is  10  feet 
in  lenjjth  makes  20  vibrations  in  35  seconds. 


548  FIRST  COURSE   IN  ALGEBRA 

3.  If  a  certain  pendulum  vibrates  once  in  a  second,  find  the  time  required 
for  a  pendulum  which  is  twice  as  long  to  vibrate  once. 

4.  Find  the  length  of  a  pendulum  which  makes  80  vibrations  per  minute 
at  a  place  at  which  the  value  of  g  is  32.16. 

5.  Find  the  length  of  a  pendulum  which  vibrates  once  per  second  at  a 
place  at  which  the  value  of  g  is  32.19. 

6.  If  at  a  certiiin  place.a  pendulum  39  inches  in  length  vibrates  once  in  a 
second,  find  the  length  of  a  pendulum  which  at  the  same  place  will  make 
one  vibration  in  one  minute. 

7.  If  a  ball  suspended  by  a  fine  wire  makes  88  vibrations  in  15  minutes, 
find  the  length  of  the  wire. 

8.  If  a  pendulum  which  is  39.1  inches  in  length  vibrates  once  in  a 
second  at  a  certain  place,  find  the  length  of  a  pendulum  which  will  vibrate 
once  in  5  seconds. 

Exercise  XXIV.     7.    Review 

1.  If  a  =  1,  J  =  3,  and  c  =  2,  find  the  value  of 

(a  +  b)ib  +  c)(c  +  a)+  a"  +  b'  +  c«. 

2.  Factor  1  +  10a;-  liar*. 

3.  Factor  x^(j/  +  1)  +  f(x  +  1)  +  a;  +  3^  +  2  a;y. 

4.  Factor  «»  -  a;*  -  a(a^  -  x^  +  x(a  -  xf. 

5.  Find  the  prime  factors  of  (a  -  a^f  +  {a"  -  If  +  (1  -  a)\ 

Simplify  the  following  expressions  : 

«(--i)-('-0-    -^^^ 


(«*  -  h''){a^  -  b') 


a       b 


^'  (a«  -  6«)(a  -b)'  13.  {x-^  +  r')  -^  (a;~*  +  y\ 

27-2  .  9  X  -1-^  _  3-8 
14.  —  ^ 


27'  •  3« 

15.   Express  (—  a  —  ^)  -^  (—  ar^  —  b'^)  with  the  minimum  number 
of  minus  signs. 


REVIEW  649 


16.   Find  the  value  of 


x-\-  2m      x  —  '2m   ^       4:mn       . .  mn 

-r  -z ; 1 — 3 : — 5'  II  £c  = 


2  71  — X       2n  +  X       x^  —  4:71^  m  +  n 

17.  Show  that  (a^  -  Pf  =  a^  -  ab  +  b\  if  a  +  b  =  1. 
Simplify  each  of  the  following  expressions  : 

18.  (3  -  ^^(2  -  V^). 

19.  ab  +  Vab  +  (a  -  \/b)(Va  -  b). 

20.  (V^  -  V=^)'  +  (\/3  -  V^y. 

21.  (V5  -  'v/?  +  2)(V5  +  V7  -  2). 

22.  (\/7  +  V^  -  V^Xa/t  -  Vs  +  Vs.) 

23.  ( Vn -  a/6  +  5)( Vll  -  a/6  -  5). 

\Vb      Va)   '    \Vb       Va) 


550  FlllST  COURSE  p  ALGEBRA 


CHAPTER  XXV 

SYSTEMS  OF   SIMULTANEOUS   EQUATIONS   INVOLVING 
QUADRATIC  EQUATIONS 

Systems  of  Two  Equations  Containing  Two  Unknowns 

1.  The  most  general  form  for  an  equation  of  the  second  degree 
containing  two  unknowns,  x  and  y,  is 

ax^  +  2hxii  +  by''  +  2  gx  +  2fy  +  6  =  0, 
in  which  «,  b,  Cyf^  g^  and  h  all  represent  real  known  numbers. 

If  one  or  more  of  these  letters  be  given  the  value  zero,  the  terms 
of  which  they  are  the  coefficients  disappear,  and  we  have  special 
types  of  quadratic  equations  containing  two  unknowns,  x  and  y^ 
such  as  the  following  : 

If  /,  g,  be  zero,  we  have  ax^  +  2  lixy  +  hy^  -\-  c  =  0.  (i.) 

f^g.h,  ax^   +  hy^  +  c  =  0.  (ii  ) 

6,/,  gf,  ax2   -f  2  hxy  +  c  =  0.  (iii.) 

a,/,  5f,  2hxy-^  by^   -\  c  =  0.  (iv.) 

b,  g,  K  ax^  -{-  2fy  -\-c  =  0.  (v.) 

<hf,hy  hy^   +  2yx  +c  =  0.  (vi.) 

etc.  etc. 

2.  In  this  chapter  we  shall  obtain  the  solutions  of  certain  sys- 
tems of  simultaneous  equations,  containing  one  or  more  equations  oi 
the  second  degree,  of  the  general  t)rpes  shown  above. 

3.  In  Chapter  XVII,  we  found  that  there  exists  a  definite  set  of 
values  of  the  unknowns  which  satisfies  all  of  the  equations  of  a  given 
set  simultaneously,  provided  that  the  given  equations  are  inde- 
pendent and  consistent,  and  that  the  number  of  equations  is  equal 
to  the  number  of  different  unknowns  appearing  in  the  system. 

We  shall  find,  whenever  one  at  least  of  the  given  equations 
composing  a  system  is  of  the  second  or  higher  degree,  that  there 


SIMULTANEOUS  QUADRATICS  551 

can  be  found  more  than  one  set  of  values  which  satisfies  all  of  the 
equations  at  the  same  time. 

4.  There  are  many  systems  consisting  of  two  equations  of  the 
second  or  higher  degrees  with  reference  to  two  unknowns  which 
cannot  be  solved  by  means  of  quadratic  or  linear  equations. 

5.  Whenever  a  system  consists  of  one  equation  of  the  second 
degree  and  another  of  the  first  degree  with  reference  to  two  un- 
knowns, say  X  and  y^  the  equation  of  the  second  degree  having  the 
form  either  of  the  general  equation 

ax^  +  2  hxy  -^  Of  -h  2  gx  +  2fy  +  c  =  0, 
or  of  one  of  the  special  forms  ax"^  -{-  bi/'^  +  c  =  0  (see  §  1),  we  may 
always  make  the  solution  of  the  system  depend  upon  the  solution 
of  equations  of  either  the  second  or  of  the  first  degree. 

6.  ^rom  the  eqiMtian  of  the  first  degree,  we  7nay  ex'press  the  value 
of  one  of  the  unknowns,  say  y,  in  terms  of  the  other,  x;  on  substi- 
tuting this  expressed  value  for  the  same  unknown,  y,  wherever  it 
appears  in  the  equation  of  the  second  degree,  we  shall  derive  an 
equation  of  the  second  degree  containing  but  one  unknown,  x,  called 
the  x-eliniinant  of  the  system ;  this  x-eliminant  may  be  solved  by 
the  methods  already  sJiown  fm^  tJie  solution  of  quadratic  equations 
containing  one  unknown. 

From  the  given  system  is  thus  derived  an  equivalent  system  con- 
sisting of  the  given  equation  of  the  first  degree  with  reference  to 
both  of  the  unknowns,  and,  as  the  case  may  be,  either  the  a;-eliminant 
or  the  ?/-eliminant  of  the  system,  which  contains  but  one  of  the  given 
unknowns. 

The  number  of  solutions  of  the  particular  eliminant  employed 
depends  upon  its  degree  with  reference  to  the  unknown  contained 
in  it,  and  by  substituting  the  solutions  of  the  eliminant  separately  in 
the  remaining  original  equation  of  the  first  degree,  we  shall,  for  each 
value  of  the  unknown  substituted,  say  x,  obtain  a  corresponding  value' 
for  the  remaining  unknown,  y. 

The  number  of  sets  of  values  thus  obtained  is  the  same  as  the 
degree  of  the  "  eliminant "  equation. 

If  the  eliminant  be  of  the  second  degree,  the  number  of  solutions 
of  the  given  system  for  finite  values  of  the  unknowns  is  two ;  if  it 
be  of  the  third  degree,  three  ;  etc. 


552  FIRST  COURSE   IN  ALGEBRA 

7.  We  shall  speak  of  the  number  of  sets  of  values  which  satisfy 
all  of  the  equations  of  a  given  system  simultaneously,  as  the  Order 
of  the  System. 

8.  We  will  state,  without  proof,  the  following  Principles : 

(i.)  The  number  of  roots  of  an  integral  equation  contai?iing  one 
unknown  is  egmil  to  the  degree  of  the  equation  with  reference  to  that 
unknown,  and  these  roots  may  be  equal  or  unequaly  rational  or  irra- 
tionalf  real  or  complejr^  acco?'ding  to  circumstances. 

(ii.)  Ths  number  of  sets  of  values  which  satisfy  all  of  the  equations 
of  a  given  determinate  system  cannot  be  greater  than,  and  is  in 
general  equal  to,  the  pi'oduct  of  ths  degrees  of  the  separate  equations 
of  which  the  system  is  composed. 

E.  g.  Since  the  equations  2  a;  +  y  =  9  and  2  a;^  +  y^  =  33  are  of  the  first 
and  second  degrees  respectively,  they  may  be  spoken  of  as  constituting  a 
1-2  system  of  the  second  oixler. 

The  product  of  the  degrees,  1  and  2,  of  the  equations  leads  us  to  expect 
that  there  are  two  sets  of  values  which  satisfy  the  equations  simultaneously. 
These  sets  are  found  to  be,  a:  =  2,  y  =  5,  and  z  =  4,  y  =\. 

9.  Since  the  ar-eliminant  or  the  y-eliminant  of  a  1-8  system 
containing  two  unknowns,  x  and  y,  is  of  the  third  degree  with 
reference  to  either  x  or  y,  it  follows  that,  to  solve  a  system  of  the 
third  order,  we  must  solve  an  equation  of  the  third  degree,  such  as 
ax^  -{■  bx^  -{-  ex  +  d  =  Oj  containing  one  unknown.  Also,  since  the 
ehminant  of  a  2-2  system  is  of  the  fourth  degree,  it  follows  that 
to  solve  a  system  of  the  fourth  order,  we  must  solve  an  equation 
of  the  fourth  degree,  such  as  ax^  +  bx^  +  ca^  +  dx  +  e  =  0,  contain- 
ing one  unknown. 

10.  Except  in  certain  very  special  cases,  the  solutions  of  systems 
of  the  third  or  higher  orders  cannot  be  made  to  depend  upon  the 
solutions  of  equations  of  either  the  second  or  of  the  first  degree. 

11.  It  should  be  observed  that,  although  the  solution  of  a  given 
system  in  which  quadratic  or  higher  equations  appear  leads  in 
general  to  the  solution  of  an  equation  of  higher  degree  than  the 
second  (that  is,  to  the  solution  of  the  eliminant  equation),  the  solu- 
tions of  a  great  number  of  systems  of  simultaneous  equations  may 
be  made  to  depend  upon  the  solutions  of  quadratic  equations. 

12.  We  will  now  consider  certain  methods  which  may  be  applied 


SIMULTANEOUS  QUADRATICS  553 

to  obtain  the  solutions  of  systems  of  two  simultaneous  equations 
containing  two  unknowns. 

I.    Slimination  by  Substitution 

13.  If  one  equation  of  a  system  of  two  simultaneous  equations 
is  of  the  first  degree,  and  the  other  equation  is  of  the  second  or 
higher  degree,  the  solution  of  the  system  may  in  certain  cases  be 
made  to  depend  upon  the  solution  of  a  quadratic  equation. 

Ex.  1.    Solve  the  system 

3x2 +  2/' =  84,     (1)^ 

>  I.    Given  System. 
3a: +  51/ =42.     (2)) 

Since  the  equations  are  of  the  second  and  first  degrees  respectively,  this 
may  be  classed  as  a  2-1  system,  and  accordingly  we  may  expect  to  find  2  •  1 
or  2  sets  of  values  for  x  and  y  which  satisfy  the  equations  simultaneously. 

From  equation  (2)  we  may  derive  an  equivalent  equation  in  which  the 
value  of  y  is  expressed  in  terms  of  x  as  follows : 

42 -3  a; 

Substituting  this  expressed  value  in  place  of  y  in  the  first  equation,  we 
obtain  the  aj-eUminant, 

3..+  (l^^)^84.  (4) 

Or,  (x  +  l)(a;  -  4)  =  0.  (5) 

The  original  system  is  equivalent  to  the  derived  system 


y  = >     (3)  /         Equivalent 

r     *       Derived  System. 
(x  +  l)(a:  -  4)  =  0.      (5)  ) 

The  solutions  of  the  x-eliminant  (5)  are 

X  =  —  1 ,     and    X  =  4. 
Substituting  these  values  for  x  in  the  remaining  equation  (3)  of  System 
II.,  we  obtain: 

Substituting  -  1  for  x,  Substituting  4  for  x, 


42  +  3  42-12 

y=9.  2/  =  6. 


654 


FIRST  COURSE  IN   ALGEBRA 


=  4,  )        These  two  groups  of  equations 
=  6.  )      form  an  eiiuivalent  System  III. 


The  two  solntions  of  the  given  equations  are  thus  found  to  1)6 

y  =      9,  j  y 

By  substitution  we  find  that  the  given  equations  are  satisfied  by  these 
sets  of  values. 

Graphical  Interpretation. 

14.    The  algebraic  problem  of  finding  the  common  solutions  of  a 
system  of  equations  containing  two  unknowns  suggests  the  graphical 

problem  of  finding  the  points  of 
intersection  of  the  graphs  repre- 
senting the  given  equations. 

In  Fig.  1  tlie  straight  line  AB, 
which  is  a  portion  of  the  graph  of 
the  linear  e([uation  3  ar  +  5  y  =  42 
(see  Ex.  1,  §  13),  cuts  the  ellipse 
which  is  the  graph  of  the  quadratic 
equation  3x^-\-  if  =  84  (see  Ex.  1, 
§  13),  in  two  points,  A  and  B. 

By  writing  the  first  member  of  the 
a:-eliminant  (5)  et^ual  to  y,  we  obtain 
the  equation  of  which  the  graph  is 
the  parabola,  a  portion  of  which  is 
shown  in  dotted  line. 

The    distinction  between   the 
solution  of  a  single  equation  con- 
taining one   unknown,   and  the 
solution  of  a  system  consisting  bf 
several  equations  containing  sev- 
eral unknowns,  should  be  care- 
fully noted. 
The  solutions  of  a  single  equation  containing  one  unknown,  that 
is,  the  values  of  its  roots,  may  be  taken  as  locating  the  points  at 
which  the  graph  of  the  equation  crosses  the  axis  of  X. 

Referring  to  Fig.  1,  it  may  be  seen  that  the  solutions  of  the  given  System 
I.  or  of  the  equivalent  derived  System  III., 


\i   ! 

Y 

1 

; 

1\U, 

— ku- 

~~~~~pff^ 

\ 

y^jLjR 

^^A[^  / 

h\\ 

"Hi^ff^" 

% 

•A 

•1 

i 

.    .            \ 

f 

0^ 

ID 

Jt 

\     '1 

/I/ 

\! 

w 

V 

y\ 

1 
1 

FlGl. 


=  — 1)  X  =  4    ) 

^'  >-    and  _'  ^    if  taken  as  coordinates, 


SIMULTANEOUS  QUADRATICS  555 

serve  to  locate  the  intersections  A  and  B  of  tlie  ellipse  and  the  oblique 
straight  line  which  are  the  graphs  of  the  given  e(iuations. 

By  eliminating  y  from  the  given  equations  (1)  and  (2)  we  derive  the  x- 
eliminant  (5),  which  is  a  conditional  equation  expressing  tlie  relations 
existing  between  the  x-values  sought,  without  for  the  moment  referring  to 
the  corresponding  t/- values. 

In  Fig.  1  it  will  be  seen  that  the  result  of  this  elimination  graphically 
is  to  locate,  by  means  of  the  intersections  of  the  parabola  with  the  axis  of 
X,  two  points  C  and  D  the  distances  of  which  from  the  origin  0  are  equal 
to  the  a:-co()rdinates  of  the  points  A  and  B  respectively. 

Tiie  roots  of  the  z-eliminant,  therefore,  gave  us  these  values, 

X  =  —  1,  X  =  4. 

The  ?/-eliminant,  y^  —  15i/  +  54  =  0  (not  shown  in  the  figure),  would 
in  a  corresponding  way  express  the  relations  existing  between  the  7/-values 
sought.  Hence  the  roots  of  the  y-eliminant,  y^  —  Wy  -{-  54  =  0,  would  thus 
give  these  values,  y  =  9  and  y  =  G. 

If  instead  of  substituting  the  values  a:  =  —  1  and  ic  =  4,  obtained  from 
the  solution  of  the  aj-eliminant  x^  — 3a:  — 4  =  0  (5)  in  the  given  linear 
equation  3a;  +  5y  =  42  (2),  we  had  substituted  these  same  values  in  the 
([uadratic  ecpuition  3x^  +  y^=S4  (1),  we  would  have  obtained,  corre- 
sponding to  each  value  of  x  substituted,  two  values  for  y  instead  of  one  as 
before.  Hence  in  addition  to  the  values  previously  found  we  would  have 
obtained  the  extra  sets  of  values 


;=:«:}  •-  ;=J.} 


By  examining  Fig.  1,  it  appears  that  these  values  serve  to  locate  extra 
points  on  the  ellipse,  marked  •JSJ.g  and  E^^,  which  do  not  lie  also  upon  the 
given  straight  line  AB. 

Accordingly  these  sets  of  values  cannot  be  accepted  as  being  solutions 
of  the  given  system. 

By  substitution  it  will  be  found  that  these  sets  of  values  do  not  satisfy 
the  given  linear  equation  (2). 

Hence  the  system  composed  of  the  solutions  of  the  x-eliniinant  and  the 
given  quadratic  ecjuation  is  not  equivalent  to  the  original  system. 

Similarly,  the  system  composed  of  the  solutions  of  the  i/-eliminant 
7/2  _  15?/  +  54  =  0  and  the  given  quadratic  equation  is  not  ec^uivalent  to 
the  original  system,  for  the  solutions  of  this  system  bring  in  the  extra 
solutions 

'  y      and  ,.    y 

y  =  9,  )  y=      0.  i 


556  FIRST  COURSE  IN  ALGEBRA 

These  extra  sets  of  values  may  be  taken  as  locating  the  extra  points 
E^  and  E_^  which  lie  neither  upon  the  oblique  line  AB  which  is  the  graph 
of  the  given  linear  ecjuation  3x  +  5y  =  42,  nor  upon  the  ellipse  which  is 
the  graph  of  the  quadratic  equation  Sx^  +  y^  =  84. 

Acconlingly  these  sets  of  values  luust  be  rejected  as  not  being  solutions 
of  the  given  system. 

The  vertical  dotted  lines  which  are  the  graphs  of  the  root  values  x  =  —  I 
and  X  =  4  of  the  x-eliminant,  by  their  intersections  with  the  oblique  straight 
line,  locate  the  points  A  and  B  respectively  of  the  line  ABj  and  no  other 
points. 

The  horizontal  dotted  lines  which  are  the  graphs  of  the  root  values 
y  =  9  and  y  =  6  of  the  y-eliminant  y'^  —  15y  +  54  =  0,  by  their 
intersections  with  the  oblique  straight  line  also  locate  the  points  A 
and  B,  and  no  other  points. 

It  follows  that  the  derived  si/stem  composed  of  the  solutions  of  either 
the  o'-eliminant  or  the  y-eliminanty  if  taken  together  with  the  given 
equation  of  the  first  degree^  is  eiiuivalent  to  the  given  system. 

Furthermore,  it  will  lie  seen,  by  referring  to  Fig.  1,  that  the  horizontal 
and  vertical  dotted  lines  intersect  in  the  four  points  A,  B,  R  and  R'. 

The  sets  of  values  x  =  —  1,  2/  =  6,  and  x  =  4,  y  =  9,  corresponding  to  the 
coordinates  of  the  points  R  and  R',  are  not  solutions  of  the  given  system, 
and  must  accordingly  be  rejected. 

Thus,  it  may  be  seen  that  to  solve  a  given  system  it  is  not  sufficient 
dimply  to  obtain  a  certain  number  of  different  valuss^  as  when 
solving  a  single  equation^  but  it  is  necessary  also  to  arrange  properly 
in  sets  the  different  values  founds  in  such  a  way  that  on  substitution 
the  values  of  each  set  will  satisfy  all  of  the  equations  of  a  given 


It  will  be  observed  that  the  dotted  straight  lines  which  are  the  graphs  of 
jc  =:  —  1, 1/  =  9,  intersect  at  -4,  while  the  dotted  lines  which  are  the  graphs  of 
a;  =  4,  y  =  6,  pass  through  B. 

From  the  reasoning  above,  it  appears  that  the  two  systems  of  equations 

x  =  -l,>  ,         a:  =  4,) 

y^      9,1         ^"^        y^%\ 

taken  together,  are  equivalent  to  the  given  S^'^stem  I. 


SIMULTANEOUS   QUADRATICS  657 

Ex.  2.    Solve  the  system 

3x^-2xy-y^  =64,     (1) )  ^     ^.         o     . 

l-ly=    2.     (2)r-    G-- System. 

Since  this  is  a  2-1  system,  we  may  expect  to  find  two  sets  of  values 
for  X  and  y. 

Substituting  in  the  first  equation  the  expressed  value  of  x  in  terms  of  y 
from  the  second  equation,  we  obtain  as  the  ^-eliminant, 

3(2  +  3yy-  2(2  +  3  y)y  -  y^  =  64.  (3) 

Or,  (y-lX5y  +  l3)=0.  (4) 

By  the  principles  of  equivalence,  the  system  composed  of  the  solutions 
of  this  eliminant,  taken  together  with  the  linear  equation  (2),  is  equivalent 
to  the  given  system. 

That  is,  (2/  -  1)(5  ?/  +  13)  =  0,     (4)  |         Equivalent 

X  —  liy  =  2.     (2)  I     '        Derived  System. 

The  solutions  of  equation  (4)  are 

y=l,         and         y  =  -^- 

Substituting  these  values  in  the  remaining  equation  of  System  II.,  we 
obtain  : 

Substituting  1  for  y.  Substituting  —  J/  for  y, 

a:  -  3  •  1  =  2  a;  -  3(-  Y)  =  2 

x  =  b.  x  =  -^' 

The  following  sets  of  values  are  solutions  of  the  given  system  of  equa- 
tions, and  by  substitution  are  found  to  satisfy  both  equations : 

^  =  M  and        ^  =  -X' 

2/  =  1,  )  2/  =  -  ¥• 

Exercise  XXV.     1 

Solve  each  of  the  following  systems  of  equations,  rejecting  all  sets 
of  values  which  do  not  satisfy  both  of  the  given  equations : 

1.  35^  +  /  =  200,  4.         x^J  =  104. 

X=  1  1/.  X  —  2/  =        5. 

2.  X  —  1/  =  —    2f  5.  icj/  =  14, 
a^  +  f=      10.                                4iB-^  =  26. 

3.  x^-^  f  =  34,  6.    x^-f  =  40, 
X  +2i/=l3.  2x  +  t/=^n. 


558  FIRST  COURSE  IN  ALGEBRA 


7. 

a-2  4-y2=  17, 
jc  — 3y=    1. 

18. 

^+    -^  =  2, 
y        X 

8. 

x'+  4/ =  32, 

Qx—  [>y=  1. 

9. 

5x  +  Qi/  =    8. 
ar»+2/  =  73, 

19. 

1          1           1 
a         y~      2' 
a;-3y  =  -  1. 

10. 

Sx  -    y  =    S. 

5x^  —  xy  =  15, 
2x  +  3y  =  36. 

20. 

X     y  __5 
y      X      2 
x-y  =  -2. 

11. 

x-h  8y  =  xy, 

21. 

4  +  5-^' 

X—    y  =  ij. 

12. 

x^  +  xy-\-     y^  = 

7, 

4      5       6 

X     y"  'o' 

X  -{■  4y  =  — 

1. 

22. 

x-y-^  =  o, 

13. 

x—  ity  = 
Sx  —  2y  +  4/  = 

h 
9. 

1  +  1-5  =  0. 
X      y 

14. 

x  +  y  +  2xy  =  Ys 

23. 

M-' 

15. 

5x-    2y  =  ^. 
2x^+  3iCj/  +  4:f  = 

=  64, 

y       2            3 

16. 

x+     y  = 

2x-Sy  = 
^x^-3xy+     y'  = 

--  —  2. 

--    2, 
I  44. 

24. 

a;4-  1       6 

y+l""5' 

a;2  +  ?/      65 

a;  +  /  ~  46 

17. 

xy  =  45. 

25. 

a;       1       7/     _61 
ic  +  ^      X  —  y      11 

2a;  +  8?/ =  54. 

II.    Reduction  of  Systems  of  Equations  by  Factoring 

15.  We  have  seen  that,  if  the  factors  of  the  first  member  of  an 
equation,  the  second  member  of  which  is  zero,  be  separately  equated  to 
zero,  the  system  composed  of  the  entire  group  of  equations  thus  formed 
is  equivalent  to  the  given  single  equation.     (See  Chap.  XII.  §  48.) 


SIMULTANEOUS  QUADRATICS  559 

E.  g.  If  a  given  equation  be  represented  by  A  •  B  -  C  =  0,  in  which 
A,  Bf  and  G  are  rational  and  integral  with  reference  to  certain  unknowns, 
then  the  system  composed  of  the  separate  equations  A  =  0,  B  =  0,  and  G=0 
is  equivalent  to  the  given  single  equation  A  •  B  '  C  =  0. 

16.  From  this  principle  it  follows  that,  if  a  single  determinate 
system  of  equations  he  represented  by  A  -  B  '  C  =  0  and  D  =  0,  in 
which  A,  B,  (7,  and  D  represent  expressions  which  are  rational  and 
integral  with  reference  to  certain  unknowns^  the  single  system 

_    '  >•  1.   (jiven  System. 

is  equivalent  to  the  group  of  separate  derived  systems 

A=Q,\r^         B  =  0,lr\        C=Q,}r"\ 

i)=o:l^''^   i>=o:p")  i>=o:[^^"-> 

(The  following  proof  may  be  omitted  when  the  chapter  is  read  for  the  first  time.) 

For,  any  solution  of  the  given  system  must  reduce  D  to  zero  and  also 
ivduce  to  zero  either  one  or  all  of  the  factors  A,  B,  or  C. 

Hence,  every  solution  of  the  given  system  must  be  a  solution  of  at  least 
"IK!  of  the  derived  sy.stems. 

Any  solution  of  (i.)  must  reduce  7)  to  zero  and  also  A  to  zero,  and 
accordingly  must  reduce  to  zero  the  product  A  •  B  '  G. 

Hence  every  solution  of  the  derived  system  (i.)  is  also  a  solution  of  the 
given  SystiMU  I. 

Similarly,  every  solution  of  any  one  of  the  derived  systems  (i.),  (ii.)  or 
(iii.)  is  also  a  solution  of  the  given  System  I. 

Accordingly  the  original  single  system  is  equivalent  to  the  group  of 
derived  systems. 

17.  The  application  of  this  principle  is  not  affected  by  the  number 
of  factors  in  the  first  member  of  any  particular  equation  the  second 
member  of  which  is  zero,  or  by  the  number  of  such  equations. 

E.  g.    Let  it  be  required  to  separate  the  single  system  of  equations 

A  ■  B  =  (\  7  ^     (.  .^gj^  System. 

a-/>  =  o,  >  ■ 

into  a  group  of  separate  systems  of  equations  which,  taken  together,  are 
equivalent  to  the  given  system. 

The  two  following  derived  systems  of  equations  are  equivalent  to  the 

given  system : 

^  =  0,  ^  /?  =  0,  > 

G-D  =  i\S  G- !)  =  ().) 


560 


FIRST  COURSE  IN  ALGEBRA 


further  separated,  we  shall 


Since  each  of  these  derived  systems  can  be 
obtain  finally 

^  =  H(i.)    ^  =  ^'l(ii.)  ^'-^4  (ill.)  ^  =  ^'l(iv.) 

These  systems  of  equations,  taken  together,  are  equivalent  to  the  given 
system  of  equations. 

Ex.  .1 .    Solve  the  system  of  equations 

2x^  +  3Ty  +  y^=    0,     (1)  >  j      (.j^.^  g    ^^m. 
a:2_3a:=10.     (2)1  '^• 

Since  this  is  a  2-2  system,  we  shall  expect  to  obtain  2  •  2  or  4  solutions. 
Writing  the  given  equations  so  that  their  second  members  shall  be  zero, 
and  factoring  the  resulting  first  members,  we  obtain  the  equivalent  system, 
(2  a:  +  y)  (x -\- y)  =  0,     (3)  >  j j      Equivalent 
(x  +  2)(x  -  5)  =  0.     (4)  )      '         Derived  System. 
By  the  principle  under  consideration,  the  given  system  is  equivalent 
to  the  foUowing  group  of  separate  derived  systems  taken  together : 

2a:  +  y  =  0,),j      2x^y  =  0,>^.^   ^  +  2/ =  «' t  (iii.)   ^  +  ^  =  ^'l  (iv.) 

We  have  thus  reduced  the  solution  of  the  given  system  of  quadratic  equa- 
tions to  the  solution  of  four  systems  of  simultaneous  linear  equations. 
The  solutions  of  thesse  separate  systems  are  found  to  be 

y=      4.  S  ^   '      »/  =  -  loJ  *■    ' 

-=-^'|(iii.)  y=  M(iv.) 

y=       2.  )  y  =  -5.> 

By  substitution,  each  of  these  sets  of 
values  is  found  to  be  a  solution  of  the 
the  given  system. 

Graphical  Interpretation 

18.  In  Fig.  2  a  portion  of  the  graph 
of  the  equation  2x^  -[-  3xy  +  y^  =  0 
(See  Ex.  1,  §  17)  is  represented  by  the 
two  oblique  straight  lines  passing 
through  the  origin  0.  The  graph  of 
the  equation  x^—3x  —  10  =  y  (see  Ex. 
1,  §  17)  is  the  parabola. 

It  should  be  observed  that,  since 
x^  —  3x  —  10  =  0  contains  the  single 
unknown  x,  this  equation  may  be  treated 


SIMULTANEOUS  QUADRATICS  561 

as  the  a:-eliminant  of  the  given  system.  Hence  our  graphical  problem  be- 
comes that  of  finding  points  on  the  oblique  lines  the  a:-coordinates  of  which 
are  equal  to  the  root  values  of  the  equation  of  which  the  graph  is  the  parabola. 

Accordingly,  the  solutions  of  this  particular  system  of  equations  may  be 
taken  as  locating  points  on  the  oblique  lines,  but  not  as  locating  the  points 
of  intersection  of  the  parabola  with  the  straight  lines.  (Compare  with 
Fig.  1,  §  14.) 

Ex.  2.    Solve  the  system  of  equations 

2.^=7;,.     (2)^-    G-- System. 
Since  this  system  of  equations  is  of  the  fourth  order,  we  may  look  for  four 
solutions. 

Transposing  all  terms  to  the  first  members,  and  factoring,  we  obtain  the 
equivalent  system 

(x-y  +  5Xx-y-b)  =  0,     (3) )  jj       Equivalent 

x(2x  —  *Jy)=  0.     (4)  )       '         Derived  System. 
This  single  system  is  equivalent  to  the  entire  group  of  separate  systems  ; 

x=:0.>^^  2x-1y  =  Q.)      ^^ 

X —  y  —  5  z=0,}  ....  .  x  —  y  —  5z=0\ 

The  solutions  of  these  systems  of  equations  are  found  to  be 

x  =  0,  >,.  ^     x  =  -7,l  ^..  .     x=      0,  }.....     a;  =  7,  )..    . 
,,  =  5:f('->    j,  =  -2.^"-)    !,  =  - 5.  ;('"•)    y  =  2.]("-^ 
By  substitution,  these  sets  of  values  are  all  found  to  satisfy  the  given 
equations. 

Exercise  XXV.     2 

Reduce  each  of  the  following  systems  of  equations  to  equivalent 
groups  of  separate  systems  of  equations,  and  solve  : 

1.  (x  -  iy)(i/  -  9)  =    0, 

x  +  i/=10. 

2.  5x^-xi/  =  0, 

4a;  —  7/  =  1. 

3.  (x  —  2)(x  +  i/-S)  =  0, 
(x-'1J-4.)0j-5)  =  O, 

4.  (x  -  y){x  +  7/  -  1)  =  0, 

{x  +  \){x  +  2)  =  0. 

5.  a;'+  ?>xy=  18/, 

ic  +  ?/  =    2. 

36 


6. 

a;2  +  3/=:12, 

x'^  —  '2xy=    3  2/1 

7. 

(x-yy-^=    0, 

(x  +  yY  =  25. 

8. 

x'  —  ^xy^  15/, 

ic2_^2^  1^    2aj. 

9. 

xy  —  &y-[-  5a;  =  30, 

x-\-y=    9. 

0. 

xy  +  x  —  y  =  25, 

x(x-y)=    0. 

662  FIRST  COURSE   IN   ALGEBRA 

11.  Q?  —  xy^  66,       13.  x^  -^-xy  ^-x  —  y=  72, 
5a;^— 16a^+  11/=    0.  '6x^  -'lxy  —  f=    0. 

12.  a^  —  /       =x  —  y,  U.  Ax^  —  ^xy -\- bx  —  y=ll, 
Q?  —  Axy  =  Ax—l&y.  x^  +  xy=    0. 

III.     Systems  of  Two  Homogeneous  Equations  of  the 

Second  Degree  Containing  Two 

Unknowns 

19.  An  equation,  one  member  of  which  is  a  homogeneous  function 
of  x  and  y  and  the  other  member  of  which  is  either  a  homogeneous 
function  of  x  and  //  or  a  kno^vn  number,  is  said  to  be  homogeneous 
with  respect  to  the  unknowns,  x  and  j/,  appearing  in  it. 

E.g.  x'^  +  xy=y\  x*  +  a:2y  +  X//2  +  2/8  =  0, 

x2  _  Axil  +  3  J/3  =  r)x  -  y,  a:2  +  a:y  +  6 1/2  =  8. 

20.  If  the  equations  of  which  a  system  is  composed  are  homo- 
geneous with  respect  to  the  unknowns  appearing  in  them,  the 
system  is  called  a  homogeneous  system. 

21.  The  solutions  of  every  system  of  two  equations  of  the  second 
degree,  which  are  homogeneous  with  reference  to  two  unknowns, 
can  be  obtained. 

22.  If  the  first  members  of  the  equations  of  a  homogeneous 
system  containing  two  unknowns,  x  and  ?/,  are  of  the  second  degree, 
while  the  second  members  are  either  known  numbers  or  homogen- 
eous functions  of  the  same  degree  with  reference  to  the  unknowns, 
and  if  these  second  members  differ  only  by  a  numerical  factor,  we 
may  obtain  the  solution  by  factoring. 

Solution  by  Factoring 

"We  may  represent  two  equations  the  first  members  of  which  are 
homogeneous  with  reference  to  two  unknowns,  x  and  y,  and  the 
second  members  of  which  are  known  numbers,  by 

aix^  -f  hixy  +  Ci/  =  di,  (1) 

a^x^  +  h^xy  +  c^y^  =  d^.  (2) 

In  these  equations  «i,  «2,  ^i,  ^2,  etc.,  represent  different  real 
known  numbers. 

The  known  terms  d^  and  c?2  may  be  eliminated  from  the  two 
equations  as  follows  : 


SIMULTANEOUS  QUADRATICS  563 

Multiplying  all  of  the  terms  of  the  first  equation  by  d^,  and  those 
of  the  second  equation  by  c?i,  we  obtain  the  equivalent  equations 

aid^Q?  4-  hxd^xy  +  Cxdiif  =  d^d^^  (3) 

a^dx^  -V  h^dixy  +  c^dxy'^  =  d^di.  (4) 

By  subtraction, 

(«iG^2  —  (tidijx^  +  (bidi  —  hidi)x7j  +  (cic?2  —  c^d-^y'^  =  0.      (5) 
Representing  the  known  expressions  in  the  different  parentheses 
by  the  letters  a,  ft,  and  c  respectively,  it  appears  that  the  derived 
equation  (5)  has  the  form 

ax^  +  hxy  +  cy'^  =  0.  (6) 

Equation  (6),  taken  with  either  of  the  original  equations  (1)  or 
(2),  forms  a  system  of  equations  which  is  equivalent  to  the  given 
system.  The  factors  of  the  first  member  of  equation  (6)  may  be 
obtained  either  by  inspection  or  by  applying  the  general  quadratic 
formula  with  respect  to  either  a;  or  ?/  as  an  unknown.  Then  the 
solutions  of  the  derived  system  of  equations  may  be  obtained  by  the 
method  of  factoring. 

Ex.  1.    Solve  the  system  of  homogeneous  equations 

2x2  -  xy  +  5  yi  =  20,      (1)  )  T       /^-         a     * 

-^         •'  \x  M-     Given  System. 

Since  the  given  system  of  equations  is  of  the  fourth  order,  we  may  expect 
to  find  four  solutions. 

In  preparation  for  the  elimination  of  the  known  terms  20  and  15,  we 
may  derive  the  equivalent  system 

6  a;2  -  3  xj/  +  15  y^  =  20  •  3,     (3)  ]  ^ ,       Equivalent 
4  a;2  +  4  a:?/  +  12  2/2  =  15  .  4.     (4)  j       *  Derived  System. 

Subtracting  the  members  of  equation  (4)  from  the  corresponding  mem- 
])ers  of  ecpiation  (3)  we  obtain, 

2  a:2  -  7  xy  +  3  y^  =  0.  (5) 

Or  (2  x-y)(x-'Sy)  =  0.  (6) 

Since  the  multipliers  3  and  4,  nsed  in  the  derivation  of  equations  (3)  and 
(4),  are  different  from  zero,  it  follows  that  ec^uation  (6),  taken  with  either 
of  the  given  equations  (1)  or  (2),  forms  a  system  of  equations  which  is 
equivalent  to  the  given  system. 

As  an  equivalent  system,  we  may  take 

x2  ^  xy  +  3  ?/2  =  15,     (2)  )  jjj       Equivalent 
(2x-yXx-'Sy)  =    0.     ((>)  )        '  Derived  System 


564 


FIRST  COURSE   IN  ALGEBRA 


This  system  may  be  resolved  into  the. two  separate  derived  systems 
which,  taken  together,  are  equivalent  to  System  III. 


2x-y=    0.  > 


x-Zy 


5,? 
0.) 


The  solutions  of  these  systems  of  equations  are 

By  substitution,  these  four  sets  of  values  are  all  found  to  satisfy  both  of 
the  given  equations. 

Ex.  2.   Solve  the  system  of  equations 

o.      ^^'^I  =  -!"'P,Ul.   GivenSystem. 
2x^-\-xy-Qy^=      4x.  (2))  ^ 

Since  this  is  a  2-2  system,  we  may  look  for  four  solutions. 

Observe  that  the  e(iuations  contain  no  known  terms,  and  that  the  terms 

—  7  X  and  4x,  which  are  the  only  ones  below  the  second  degree,  are  similar, 

—  that  is,  differ  only  by  a  numerical  factor.     Hence  we  may  eliminate,  these 
terms  and  solve  the  system  of  equations  by  the  method  of  factoring. 

One  solution  of  the  system  may  be  obUiined  inmiediately  by  inspection. 
It  may  be  seen  that  if  ?/  be  given  the  value  zero,  the  first  equation  reduces 
to  x^  +  7  a:  =  0,  and  the  second  equation  reduces  to  2  x^  —  4x  =  0.  These 
ecjuations  have  in  common  the  solution  x  =  0,  and  no  other.  Hence,  to 
the  value  ?/  =  0,  in  either  equation,  corresponds  the  single  value  aj  =  0. 

Accordingly,  one  solution   of  the 

given  system  of  equations  is      _  r.  t" 

The  remaining  solutions  of  the 
system  may  be  obtained  as  follows  by 
the  method  of  factoring : 

By  eliminating  the  terms  of  the  first 
degree  from  equations  (1)  and  (2)  we 
obtain  the  homogeneous  equation 
18x2  +  7  XT/  -  30  i/2  =  0.  (3) 

Equation  (3),  taken  together  with  either  of  the  given  equations,  consti- 
tutes a  system  equivalent  to  the  given  system. 
Accordingly  we  have 

18x2 +  7x?/- 301/2  =  0, 


Fig.  3. 


.2  +  37/2     =-7x.     (I)j 


(.3)  \         Equivalent 


Derived  System. 


SIMULTANEOUS   QUADRATICS  565 

Whenever  the  factors  of  an  expression  such  as  the  first  member  of  equation 
(3)  are  not  readily  obtained  by  inspection,  we  may  proceed  as  follows  : 

Solving  (3)  as  a  quadratic  with  reference  to  x,  y  being  for  the  moment 
taken  as  a  known  number,  we  have 


^  _  -  7 1/  ±  V49  j/'^  -  4(18)(-  30  y^) 

That  is,  a:  =  i^,     (G)         and        x  =  -^'  (7) 

The  factors  of  the  first  member  of  equation  (3)  may  immediately  be 

written  from  (6)  and  (7)  which  are  the  expressed  values  of  x  in  terms  of  y. 

Hence  (3)  becomes     (9 x  -  10 y)(2  x  +  3  y)  =  0.  (8) 

Accordingly  the  derived  System  II.  may  be  written  in  the  equivalent 

form, 

(9  x  -  10  ?/)  (2  a;  +  3  iy)  =  0,  (^)  I  t  t  t      Equivalent 

x^  +  '3y^  =  -7x.     (1)1        '         Derived  System. 
The  student   should   complete   the   process  and   obtain   the   remaining 
solutions  of  the  given  system.     (See  Fig.  3.) 

Exercise  XXV.     3 
Solve  the  following  systems  of  homogeneous  equations  : 

1.  x'  +  xf/    =6,  7.    (2x  +  y)(2f/  +  x)  =  500. 
xu  +i'Mf  =  8.  (x  +   y){x-  y)  =    75. 

2.  ^^+0.^+15=0,  8.    13^y-2^^-18/  =  12, 
2x^+'6xy+  y^=lO,  2x^-bxy-    ^y=U. 

9.   ()x(x-\-  2y)    +2/'     =4, 

3.  x"  +  2xy  =  39,  ^2  ^  4^(^  +  2?/)  =  16. 

^     ^  '  10.    ?>x{x^2y)    +5/ =  21, 

4.  x^-     ,/=    3,  x'^-2y(x^2y)  =  2^. 
5ar^-4a!y +  31/2=  15.          n.    (2x  +  y){:2y  +  x)  =    |, 

5.  x^^xy^f=l,^  ^"  +  ^)^"     -^^^^^- 
x'~xy  +  if=    7.                  12.   x  +  y  =  -^ 

X 

6.  5  3;=^- 18a!y  +  16/=    9,  _^=1. 
2a3'-    Sicy-    3^^=12.              ^      ^      y* 


566  FIRST  COURSE  IN  ALGEBRA" 

Solution  by  Expressini?  the  Value  of  One  Unknown 
as  a  Multiple  of  the  Other 

23.  The  solution  of  a  system  of  two  equations  which  are  of  the 
second  degree  and  homogeneous  with  reference  to  two  unknowns, 
X  and  y^  may  be  obtained  by  expressing  the  value  of  one  unknown, 
as  a  multiple  of  the  other. 

24.  If  by  letting  x  =  ??y,  we  express  the  valvs  of  one  of  the  un- 
knownSy  Xj  as  a  multiple  of  the  other^  y^  we  may  substitute  vy  for  x 
and  obtain  a  derived  system  of  equations  in  which  v  and  y  are  to  be 
regarded  a^  the  unknowns. 

By  eliminating  y  from  the  new  system  of  equations  thus  obtained^ 
we  shall  obtain  a  single  equation  in  which  v  is  the  only  unknown. 
This  equation  is  the  v-eliminant  of  the  new  system. 

Ths  system  of  three  equations  consisting  of  the  v-eliminant^  the 
assumed  equation  x  =  vy,  and  either  of  the  equations  obtained  by 
substituting  ly  for  x  in  one  of  the  oi'iginal  eqiujitions,  institutes  a 
system  of  equations  which  is  equivalent  to  the  given  system  of  two 
equations. 

The  values  of  v  found  by  solving  the  v-eliminant^  when  substituted 
in  the  remaining  equations  of  the  new  system^  determine  the  values  of 
the  unknowns'  x  and  y. 

25.  It  should  be  observed  that,  on  condition  that  y  is  different 

X 

from  zero,  we  may  from  x  =  vy  obtain  -  =  v. 

Hence,  for  finite  values  of  x,  v  increases  in  value  indefinitely,  — 
that  is,  it  becomes  infinite  when  y  diminishes  in  value  numerically, 
that  is,  when  y  s^pproaches  zero.  Hence,  whenever  x  is  replaced  by 
vy,  the  solutions  of  which  y  =  0  is  a  part  musty  if  they  exists  be 
obtained  separately. 

Ex.  1.    Solve  the  system  of  homogeneous  equations 

.^+2,^^17,     (1)>    ^     Given  System. 
xy-     2/2=    2.     (2))  ^ 

Since  this  system  is  of  the  fourth  order,  we  may  expect  to  find  four 
solutions. 

In  this  particular  example  it  will  be  noted  that  if  \j  is  given  the  value 
zero  the  resulting  equations  have  no  common  solution  with  reference  to  x. 
Hence  y  =  0  is  no  part  of  a  solution  of  the  given  system. 


SIMULTANEOUS   QUADRATICS  56T 

If  we  assume  that  x  =  vy,  we  may,  by  substituting  vy  for  x  in  the  given 
equations,  obtain  the  equivalent  system 

(i;t/)2  +  2w2=  17,     (3)1  _     .     , 

/\  2        o      ),i\  Tr    Eciuivalent 

My  -  y^  =  2,    (4)  III.     \^   .    ,  „   ^ 

;_.  Derived  System. 

X  =  vy.      (5)J  ^ 

To  obtain  the  solutions  of  System  II.,  we  may  proceed  as  follows: 
From  equation  (3),  From  equation  (4), 

y^  =  -T^-      (6)  y^  =  -^.      (7) 

Since  the  given  equations  are  simultaneous,  these  "expressed"  values  for 
y^  must  be  equal. 

That  is,  ^Z_^=:^_^.  .     (8) 

Or,  2v2_i7v  +  21  =  0.  (9) 

The  solutions  of  the  v-eliminant  (9)  of  the  derived  System  II.  are  found 
to  be  V  =  I    and     V  =  7.  (10) 

Since  neither  of  these  values  could  have  been  introduced  by  the  multi- 
pliers -u^  +  2  and  v  —  Ij  when  deriving  (9)  from  (8)  they  must  also  be  roots 
of  equation  (8). 

The  system  composed  of  the  -y-eliminant  (9),  the  assumed  equation 
X  =  vy  and  either  equation  (6)  or  equation  (7),  constitutes  a  system  of 
equations  which  is  equivalent  to  System  II. 

2u2_i7^  +  21=     0,       (9)  \ 

2  _      2  (  jj J      Equivalent 

^  ~  1)  —  V  (       '        Derived  System. 

x  =  vy.         (5)  / 
Substituting  the  solutions  of  equation  (9),  v  =  3/2  and  v  =7,  in  equa- 
tion (7),  we  olitain : 

Substituting  3/2  for  v,  Substituting  7  Jbr  v, 

2/  =  ±  2.  y  =  ±  i  V3. 

To  find  the  corresponding  values  of  x,  we  may  either  substitute  these 
values  of  y  in  the  given  equations  (1)  or  (2),  or  we  may  substitute  corre- 
sponding values  of  v  and  y  in  x  =  vy. 


Substituting 

Substituting 

Substituting 

Substituting 

ty  =  2. 

(y  =  -2. 

(^-7, 

(.  =  7, 

( y  =  i  V3- 

\y  =  -  i  V3. 

We  find  that 

We  find  that 

^=(1)2      ■ 

x  =  ^(-2) 

X  =  7aV3) 

x  =  u-  jv^) 

x^3. 

x  =  -d. 

X  =  W^' 

x  =  -W-6. 

568 


FIRST  COURSE   IN   ALGEBRA 


The  four  solutions  of  fche  given  system  are  thus  Ibiuid  to  be 


x  =  -3. 


.  =  -2.[(«-> 


y  =  -  i  V3.  ) 


y  =  W^' 

These  four  sets  of  values  are  found  to  Siitisfy  the  equations  of  the  given 
system. 

By  extracting  the  squai-e  root  of  3,  we  may  obtain  from  (iii.)  and  (iv.) 
approxiiuiite  values  of  x  and  y,  correct  to  any  required  number  of  significant 
figures. 

Ex.  2.   Solve  the  system  of  homogeneous  equations 

x«  +  xy-y2=      29,     (1)\  .      ^.        _,    ^ 
2x«-xy-,/  =  -19.     (2))^-     Given  System. 

Since  this  is  a  2-2  system,  we  may  expect  to  find  four  solutions. 

By  assigning  the  value  zero  to  y,  it  may  be  seen  that  the  corresponding 
values  of  x  in  the  two  equations  are  not  equal.  Accordingly,  it  follows  that 
y  =  0  is  no  part  of  any  solution  of  the  given  system. 

Hence  we  may  assume  that  x  =  vy,  and  substituting  vy  for  x  in  the  given 
equations,  we  obtain  the  equivalent  derived  system 

x=      iry.     (5)J  ^ 

Observe  that,  by  dividing  the  corres- 
ponding members  of  the  two  equations 
(3)  and  (4),  the  unknown  y^  may  be 
eliminated,  and  we  may  inmiediately  ob- 
tain the  i>eliminant  of  System  II. 

v^  -\-  V  —  I 

Hence    77  v^  _  jq ^  _  43  =  o.  (7) 

The  solutions  of  (7)  are  found  to  be 

v  =  ^,      and      v  =  —  ^.  (8) 

The  system  composed  of  the  i;-eliminant 

(7),  either  one  of  the  equations  (.3)  or  (4) 

in   System  II.,  and  the  assumed  equation 

(5),  constitutes   a  system   equivalent   to 

^iG-  4.  Svstem  II. 


77  ^2  _  10  V  -  48  =    0, 

0) 

vY  +  rf  -  f  =  29, 

(3) 

X  =  vy. 

(5) 

III. 


Equivalent 

Derived  System. 


Substitutii 


SIMULTANEOUS   QUADRATICS  569 

Substituting  the  solutions  -y  =  f  and  v  =  —  J^  of  (8)  iu  (3),  we  obtain, 
the  values  ior  y. 

Substituting  ^  for  v,  Substituting  —  ^^  for  v, 

We  find  y^  =  49.  We  find  -6y'^  =121. 

Or,  2/  =  ±  7.  Or,  y  =  ±  VV^^- 

The  values  of  x  may  be  found  by  substituting  corresponding  values  of  v 
and  y  in  the  assumed  equation  x  =  vy  (5),  as  follows  : 

ing  1"""^^^  Substituting  j  ^  =  "  i't'    

"    U  =  ±  7.  -  1  ,^  =  ±_i^V-  5. 

We  find  x  =  ±6  and  a:  =  =F  f  a/-^- 

Accordingly  the  solutions  of  the  given  system  of  equations  are  : 

-=+64(i.)  ^=-iVEZ4(iii.) 

^  =  -6'[(ii.)  ^  =  +^^:z^'l(iv.) 

By  substitution  these  values  are  all  found  to  satisfy  the  given  equations. 

Since  we  have  found  four  sets  of  values,  it  appears  that,  in  deriving  the 
-y-eliniinant  (7),  no  solutions  were  lost. 

By  referring  to  Fig.  4,  it  will  be  seen  that  the  graphs  of  equations  (1) 
and  (2)  intersect  in  but  two  points,  the  coordinates  of  which  are  ic  =  6, 
y  =  7,  and  a;  =  —  6,  y  =  —  7. 

We  nmst  accordingly  interpret  the  imaginary  values  (iii.)  and  (iv.)  of 
X  and  y  as  indicating  that  the  graphs  have  no  points  of  intersection  the  co- 
ordinates of  which  are  these  solutions. 

Exercise  XXV.    4 
Solve  the  following  systems  of  homogeneous  equations : 

1.  3a;' +  2x1/  +  3/ =  88, 
2x^  —  Sxi/+  2y^  =  S7. 

2.  x''  +  4:Xi/  +  f  =  -  11, 

7  a' -3  2/'=      51. 

3.  3x^  —  2x1/+     f  =  h 

4.  x'^-xij  +  f=      93, 

x^  +  2x1/  =  —  40. 

5.  2x1/  + f  =    51, 

Sx^—x!/=  126. 


6. 

3a;''+10a;?/4-  3/ =  -21, 

x'-    y'=        5. 

7. 

x^-  3aj^=10, 

5a^-13/=33. 

8. 

x'  +  Sxi/-f^2d, 

lx^-2f=13. 

9. 

3x(x-Stj)+f=:-lh 

x^  +  2i/(x-3tj)=21. 

10. 

2x'-ly{x-y)  =  23, 

3x{x-2y)-'oif=    3. 

570  FIRST  COURSE   IN   ALGEBRA 

IV.    Reduction    of   Systems  of  Equations   by  Division 

26.  Representing  by  A,  //,  C,  and  D  expressions  which  are  in- 
tegral with  reference  to  two  unknowns,  x  and  //,  it  may  be  seen  that 
//*  the  meryibers  A  •  C  and  BJ)  of  one  equation  (1)  of  a  JSt/stem  L, 
composed  of  two  equations,  contain  as  factm-s  the  cori-espondimj  mem- 
bers A  and  B  of  the  remaining  eqiuntion  (2)  of  the  system,  then  the 
given  Si/stem  I.  is  eguival-etit  to  the  derived  double  system  (i.)  a?id  (ii.). 

That  is,  AC=BD,     (1))      ^.        ^    ^ 

1  =  B  (2^  \     ^^^®°  System. 

is  equivalent  to  the  double  system 

It  should  be  observed  that  the  derived  equation  C  =  1)  (8)  is 
obtained  by  dividing  the  members  of  equation  (1)  by  the  corre- 
sponding members  of  (2),  while  the  remaining  ec^uation  (2)  of 
the  given  System  1.  is  carried  over  unchanged  into  the  derived 
system  (i.). 

The  remaining  system  (ii.)  is  composed  of  the  equations  formed 
by  equating  to  zero  separately  the  factors  A  and  B  which  are 
common  to  the  corresponding  members  of  equations  (1)  and  (2) 
of  the  given  system. 

The  Principle  may  be  established  as  follows  : 

Substituting  for  i?  in  (1)  the  equal  value  A,  we  obtain  the  equivalent 
equation  A  •  C  =  A  •  D. 

Or,  A-C-A'D  =  0.  (6) 

Factoring,  A{C-D)  =  0.  (7) 

Accordingly,  from  System  I.  we  may  obtain  the  equivalent  system 
A{C-D)  =  Q,     (7)  1  J  J     Equivalent 

A-B  =  0.     (8)  >      '        Derived  System. 
By  the  principle  of  §  16,  this  single  system  is  equivalent  to  the  derived 
double  system 

C-D  =  0,     (9)>      ^^^^  ^=0,  (10)  > 

A-B  =  (},     (8)  i  A-B  =  0.     (8)  ) 

These  systems  in  turn  are  equivalent  to 

C=A    (3)l(i.)       and     ^=0'     Wkii.) 
4  =  ;?,     (2)  i  ^  '^  i>'  =  0.     (5)  P     ' 


SIMULTANEOUS   QUADRATICS  571 

Ex.  I.    Solve  the  system  of  equations 

2a:2-z  =  i/2_  1,     (1))^       n-         a     ^ 

,       )J.  c  I-     Given  System. 

Since  the  members  of  equation  (2)  are  contained  as  factors  in  the  corres- 
ponding members  of  equation  (1),  the  given  single  System  I.  is  equivalent 
to  the  derived  double  system 

x  =  y+\,     (2)i''  J/  +  1=0.     (5)P    -" 

Equation  (3)  is  formed  by  dividing  the  members  of  equation  (1)  by  the 
corresponding  members  of  equation  (2).  Equations  (4)  and  (5)  are 
obtained  by  eipiating  separately  to  zero  the  members  of  equation  (2)  wliich 
aie  contained  as  factors  in  the  corresponding  members  of  equation  (1). 

The  solutions  of  the  derived  systems  (i.)  and  (ii.)  are 

a;  =  -1,7  1         x  =  0,      > 

>•         and  '      > 

The  number  of  sets  of  values  thus  obtained  is  equal  to  the  order,  2,  of 
the  given  sj'stera.  By  substittition  these  sets  of  values  will  be  found  to 
satisfy  both  of  the  given  equations. 

27.  Whenever  one  of  the  expressions  represented  hy  A  or  B  (see 
§  20)  is  a  known  number,  it  follows  that  the  derived  system  (ii.) 
A  =  0,  (4),  B  =  0,  (5),  will  have  no  finite  solutions. 

Ex.  2.    Solve  the  system  of  equations 

a:8-,/  =  26,     (1)  |  j      Given  System. 
X  -y  =    2.     (2)3 

Dividing  the  members  of  equation  (1)  by  the  corresponding  members  of 
equation  (2),  we  may  derive  the  equivalent  system 

a;2  +  xy  +  1/  =  13,     (3)  >  jj^     Equivalent 

X  —  y  =    2.     (2)  )      *         Derived  System. 

Observe  that  by  separately  equating  to  zero ,  the  factors  x  —  y  and  2, 
which  are  common  to  the  corresponding  members  of  equations  (1)  and  (2), 
we  would  have  as  one  of  the  expected  conditional  equations  a  known  num- 
ber equal  to  zero,  that  is,  2  =  0. 

Hence  the  expected  derived  "double"  system  reduces  to  a  single  Sys- 
tem II.,  equivalent  to  the  one  given. 

Accordingly,  although  the  given  system  of  equations  is  of  the  third  order, 
the  number  of  finite  solutions  does  not  exceed  the  order  of  the  derived 
equivalent  System  II.,  that  is,  there  will  be  but  two  sets  of  values. 


572  FIRST  COURSE  IN  ALGEBRA 

The  solutions  of  the  derived  System  II.  are  found  to  be 

=  ''^\         and        ^  =  -'^i 
=  1,  S  2/  =  -  3,  S 


X  =  3, 
both  of  which  satisfy  each  of  the  given  equations 


Exercise  XXV.    5 
Solve  each  of  the  following  systems  of  equations  : 

1.  jc»  =  i/(x  +  ^),  8.  x^  +  f  =  72, 
x^  =  x-\-  y.  ic  +  3/  =    6. 

2.  1+  .?/  =  a^  9.  a;«  -  /  =  7, 
1  +  /  =  a*.  x  —  y  =  2. 

3.  ic(a;  -  3)  =  4  -  y*,  10.  ic«  -  i/»  =  a»  -  ^>», 

ar  =  2  +  y-  X  —  y  =  a  —  b. 

4.  aj(y+3)==9y-l,  11.  a;«  +  /  =  91, 

a;  =  3y— 1.  ic'-a-y  H-/=  13. 

5.  x{x  -a)  =  b^-  f,  12.  a;"  -  7/8  =  19, 

X  —  a  =  b  +  y.  x^  -\-  xy  +  y'^  =  Id. 

6.  a;*^  -  /  =  77,  13.  27  «;*-/  =  0, 

x  —  y=l.  9  a;2  +  3  a://  +  /  =  243. 

7.  a;  H-  ?/  =  14,  14.  x^  +  xhf  +  /  =  21, 
ar*-.2/'  =  ^6.  x"  -  xy  +  y""  =    3. 

V.     Systems  of  Symmetric  Equations 

28.  An  equation  is  said  to  be  symmetric  with  respect  to  the 
unknowns  appearing  in  it,  when  its  members  remain  unaltered  in 
value  if  any  two  of  its  unknowns  are  interchanged. 

29.  The  necessary  and  sufficient  condition  that  an  equation  be 
symmetric  with  respect  to  two  unknowns,  x  and  y,  is  that  the 
coefficients  of  like  powers  of  the  unknowns  be  equal. 

E.  g.    The  following  equations  are  symmetric  with  respect  to  x  and  y  .- 

a;  +  2/  =    5,  x^  -  xy -\- y'^  =  1  ^ 

a;2  +  2/2  =  10,  a:V  -\-xy  +  \  =0. 


SIMULTANEOUS   QUADRATICS  573 

30.  A  system  of  two  equations  is  said  to  be  symmetric  with 
respect  to  two  unknowns,  x  and  y^  if  the  equations  obtained  by 
interchanging  x  and  y  are  identical  with  those  given. 

E.g. 


31.  It  follows  from  the  principles  of  symmetry  that  if  any  solu- 
tion of  a  system  of  two  symmetric  equations  containing  two  un- 
knowns, X  and  J/,  be  represented  by  a;  =  «,«/  =  6,  then  x  —  b^y  —  a, 
will  also  be  a  solution. 

Ex.  1.    Solve  the  system  of  symmetric  equations 

x  +  y  =  4,     (1)|^     Given  System. 

Observe  that  botli  equations  are  synmietric  with  respect  to  x  and  %  and 
that  the  first  member  of  equation  (I)  is  the  sum  of  x  and  y. 

By  combining  the  given  equations  in  such  a  way  as  to  obtain  the  differ- 
ence between  x  and  i/,  it  will  be  possible  to  make  the  solution  of  the  given 
system  of  equations  depend  upon  the  solution  of  a  system  of  two  linear 
equations. 

Squaring  the  members  of  (I),  a;^  +  2 x?/  +  i/^  =  16.     (3) 

Multiplying  members  of  (2)  by  4,  -{-4xy  =12.     (4) 

By  subtraction,  x'^  —  2xy  -^  y^  =    4.     (5) 

The  given  System  I.  may  be  replaced  by  the  following  equivalent  derived 
system  of  the  same  order : 

a;  4- 1/ =  4,     (1)?tt        Equivalent 
x^  —  2xy  -^  y^  =  4.     (5)  >  Derived  System. 

This  derived  system  is  equivalent  to 

X  -{-y  z=  4,     (1)  ?  TTT       Equivalent 
(x-yy  =  4.     (6)  >        '  Derived  System. 

System  III.  is  equivalent  to  the  set  of  two  derived  systems, 

a;-2/  =  2,  y^^  x  -  y  =  -  2A  ^ 

The  solutions  of  these  systems  are 


y  =  l,> 


and 


=   1,? 

=      3.  S 


y 

Both  of  these  sets  of  values  satisfy  each  of  the  equations  of  the  given 
system  of  the  second  order.     (See  Fig.  5.) 


574 


FIRST  COURSE  IN  ALGEBRA 


If  with  the  derived  equation  (5)  we  had  used  the  given  equation  of  the 
second  degree  (2)  instead  of  the  equation  of  lower  degree  (1)  to  form  the 

derived  system,  we  would  have  passed  Irom 
the  given  1-2  system  to  a  derived  2-2  sys- 
tem, and  accordingly  the  two  systems  would 
not  have  lieen  equivalent. 

The  derived  2-2  system  would  have  been 
found   to   contain,  besides   the  solutions  of 
the  given  system  of  the  second  order,  the 
two  additional  solutions 
x=z  —  3,y  =  —  lj   and   x  =  —  l,y  =  —  3. 

These  would  have  been  introduced  dur- 
ing the  process  of  solution  by  squaring  the 
members  of  equation  (1). 

Ex.  2.  Solve  the  system  of  symmetric 
equations 


Fig.  5. 


/  >-^x  M-   Given  System. 
x+ 2/=    1.(2)3 


Since  this  is  a  2-1  system  we  may  expect  to  obtain  two  sets  of  values 
which  satisfy  both  efjuations. 

"We  will  first  obtain  the  value  of  the  difference  x  —  y. 

Squaring  the  members  of  ef|uation  (2)  and  subtracting  from  the  corre- 
sponding members  of  equation  (1),  we  obtain  the  equation  —  2arv  =  24.  (3) 

Combining  the  corresponding  members  of  equations  (1)  and  (3)  by 
addition,  we  obtain 

a:2  _  2  arj/  -I-  2/2  =  49. 
The  given  system  is  equivalent  to  the  derived  system 
a:2-2a:y  +  7/  =  49,     (4)  >  jj    Equivalent 

as  4- 1/  =    1.     (2)  )      '       Derived  System. 

This  single  system  is  equivalent  to  the  system 

{X  -  yy  =  49,     (5)  I  jjj       Equivalent 
X  -\-  y    z=    1.     (2)  j        '         Derived  System. 

This  last  system  is  equivalent  to  the  set  of  two  systems 


(4) 


X  — 
X 


-^^  =  +;'|(i.)and  ^-^  =  -M(ii.) 

+  2/=       lA  x-^y=      1.  ) 

The  solutions  of  these  systems  are 

=  -  3,  f  2/  =      4.  i 


X  = 

y 


SIMULTANEOUS   QUADRATICS 


575 


These  sets  of  values  are  found  to  satisfy  each  of  the  given  equations. 
(See  Fig.  6.) 

Ex.  3.  Solve  the  system  of  symmetric 
equations 

a;2  +  i/2=  100,     (1)")  ^    ^. 

.o      /^x  r  I-  Given  System. 
xy=    48.     (2)j  ^ 

We  may  expect  to  obtain  four  solu- 
tions, since  this  is  a  2-2  system. 

To  obtain  the  solutions  we  will  find 
the  sum  and  also  the  difference  of  the 
unknowns,  as  follows : 

Multiplying  the  members  of  equation 
(2)  by  2,  then  combining  with  equation 
(1)  by  addition  and  subtraction  succes- 
sively, we  obtain 


a:2  +  2a:7/  +  i/=196,     (3)") 
a;2  -  2  xy  -\-  if  =      4.     (4)  j 


XL 


Fig.  6. 

Equivalent 

Derived  System. 


Equation  (3)  may  be  written  in  the  form  (x  +  y)^  -  196  =  0,  which  is 
equivalent  to  (x -\- y  -\-  14)(x-  -\- y  -  14)  =  0,  (5).     Similarly,  equation  (4) 

may  be  written  in  the  form 

(x  -  yy  -4  =  0, 
which  is  equivalent  to 

(x-y-{-2)(x-y-2)  =  0,      (6). 

Accordingly  System  II.  is  equiva- 
lent to  the  following  derived  svstem  : 


III. 


(x+y+U)(x  +  y-U)  =  i\    (5), 

(x-y+   2Xx-y-    2)=0.    (6)] 

Equivalent  Derived  System. 

System    III.   is   equivalent   to   the 
roup  of  systems  of  equations 


Fig.  7. 


-2/ =  -2.  I  ^^ 


(i-) 
=  -  14, 


a:  +  7/  =  14, 
X  —  y  =    2. 

x+?/  =  -  14,")     ...  x  +  y 

X  ~y=       2.  J  ^     '^  x  —  y  =  —    2. 

From  these  systems  of  equations  we  obtain  the  following  solutions,  which 

are  numlxjred  to  correspond  to  the  systems  from  which  they  are  obtained. 

(See  Fig.  7.) 


:! 


(iv.) 


676  FIRST  COURSE  IN  ALGEBRA 

X  =  8, )     .  ^  X  =  6, )    ...  V       a;  =  —  6, )   ,...  .    x  =  —  8,  )   ,.    ^ 

By  substitution  these  values  are  all  found  to  satisfy  the  given  equations. 

Exercise  XXV.     6 
Solve  the  following  systems  of  symmetric  equations  : 

l-^-^^'  =  "'  13.1+1=11. 

xy  =  20.  X       y 


2.  a^  +  /  =  61, 

-,  +  -2  =  73. 

x-\-y  =11. 

x^      y' 

3.  x"  +  y'  =  53, 
x  +  y  =    9. 

14. 

1  +  1  =  -    4, 

X     y 

4.  a^  +  /  =  73, 

i-  =  -45. 

a;  +y  =11. 

ajy 

5.    a;+y  =  15, 
a-y  =  5(-. 

15. 

1  +  1=    1, 

X      y 

1 

6.  a;»  +  f  =  126, 

'"^=      132- 

jc  4-  3^  =      6. 

i-^  =  -- 

1.  a^  +  xy  +  y^  = 

73, 

16. 

xi/  = 

8. 

1+1=6. 

a;       ?/ 

8.  ar^  -  a?y  +  /  = 

.^^, 

xy  = 

i. 

1        1 

9.  aj2  +  a;^  +  /  = 

61, 

17. 

x^^f  =  ''^ 

a;  + ?/  = 
10.  ar'  +  a;^^     +  y""  ■ 

9. 
=    37, 

^  =  ^- 

x  +  y  - 

=  -7. 

18. 

a;?/(a;  ■\-y)  =  30, 

11.  ar^  +  y  =  ^ 

a;  +  ?/  _    5 

X  +  y  =  a. 

a;?^             6 

12.  a;  +  ?/  =  a, 

19. 

aj*  +  a^y  +  7/  =  21, 

xy  =  b. 

X-—  xy   +/=    3. 

SIMULTANEOUS  QUADRATICS  577 

i)  =  ¥ 

xy=  1. 


13  23.  (a;+  l)(v  +  1)  =  Y 

20.  x"  +   xy  -\-y'=    —>  ^  ^V         ^ 


ic*  +  icy +  /==^ 


21.  x"  ■{- xy  -{■  y''  =  84, 
X  +  a/^  +  ^'  =  14. 

a?      1^      3 

22.  x^  +  iK//  +  /   =  133. 

jc  —  Vicy  +  y  =      7. 

a;+  1      y+  1      4 

Solutions  by  Special  Devices 

32-  Certain  systems  of  equations  are  of  such  special  forms  that 
special  methods  must  be  employed  to  obtain  their  solutions. 

Ex.1.    Solve  the  symmetric  system  of  equations 

2      Q     T  r         /^(  r  I-   G^iven  System. 
2/2  =  3 1/  +  5  X.     (2)  )  -^ 

By  addinf;  the  corresponding  members  of  equations  (1)  and  (2),  we 
obtain  x^  +  t/^  =  8  (a:  +  y),  (3)  ;  and  by  subtracting  the  members  of  equa- 
tion (2)  from  the  corresponding  members  of  equation  (I),  we  obtain 
x'  -  if  =2(y-  x),  (4). 

Hence,  the  given  system  cf  equations  is  equivalent  to  the  following 
derived  system : 

x^  +  2/2  -  8(a:  +  y),     (3)  |  Equivalent 

a;2  —  2/2  =  2(y  —  x).     (4)  )       '        Derived  System. 
Transposing  the  terms  of  equation  (4)  to  the  first  member  and  factoring, 
we  obtain  the  equivaleiit  derived  equation 

(a;  +  2/  +  2)(a;-2/)=0.  (5) 

Accordingly,  equations  (3)  and  (5)  taken  together  constitute  the  follow- 
ing system  of  equations  which  is  equivalent  to  the  given  system  : 
a;2  +  1/2  =  8(x  +  y),     (3)  )  Equivalent 

(ar  +  2/  +  2)(ic  —  2/)  =  0.     (5)  j         *       Derived  System. 
The  solutions  of  System  III.  may  be  obtained  by  the  method  of  §  16. 
Ex.  2.     Solve  the  symmetric  system  of  equations 

.y  +  .,      12=   0,     OH  J      oi,en  System. 
a;2  4-  2/2  =  10.     (2)  ) 

Solve  the  first  equation  for  xy  as  the  unknown  by  factoring.  Then, 
using  the  values  thus  found  with  equation  (2),  solve  the  two  resulting 
symmetric  systems. 

87 


578  FIRST  COURSE  IN  ALGEBRA 

Ex.  3.     Solve  the  symmetric  system  of  equations 

xy=    8.     (2)j  -^ 

By  multiplying  the  membei-s  of  ec^uation  (2)  by  2  and  adding  the  results 
thus  obtained  to  the  corresponding  members  of  equation  (1),  we  obtain  the 
derived  equation 

(x  +  yy  ^x  +  y  =  90.  (3) 

Hence  the  given  system  of  equations  is  eriuivalent  to  the  system 
(x  +  y  +  10)(a:  +  1/  -  9)  =  0,     (4)  )  jj     Equivalent 

xy  =  8.     (5)  )      '        Derived  System. 

This  system  of  equations  is  equivalent  to  the  following  set  of  two 
systems  : 

a:  +  y+10  =  0,)  ,  +  ,,_  9  =  0,1 

xy  =  H.\^''^  x./  =  8.  [("•-> 

The  solutions  of  these  systems  of  equations  may  be  obtained  by  applying 
the  method  of  §  31. 

33.  Systems  consisting  of  equations  which  become  symmetric 
by  changing  the  signs  of  one  or  more  terms  may  be  solved  by 
the  methods  employed  for  the  solution  of  systems  of  symmetric 
equations. 

Ex.  4.   Solve  the  system  of  equations 

^''-^^  .1'     i\l  \  I-  Given  Svstem. 
xy  =  16.     (2)  ) 

Multiplying  the  members  of  equation  (2)  by  4  and  adding  the  results  to 
the  squares  of  the  corresponding  members  of  equation  (1),  we  obtain  the 
equation  (x  +  yf  =  100,  (3). 

Accordingly,  the  given  system  of  equations  is  equivalent  to  the  system 
(x  +  yy  -  100  =  0,     (3)  ]  Equivalent" 

X  —  y  =  6.     (1)  I      '         Derived  System. 

This  system  of  equations  is  equivalent  to  the  following  set  of  two  systems 
of  equations,  the  solutions  of  which  should  be  obtained  by  the'  student : 
,  +  ,  +  10  =  0,|  .  +  .,_  10  =  0,1 

x-y  =  (i.\  ^^  x-y  =  6.)  ^^ 

34.  A  system  of  equations  having  the  forms 

ax  +  bi/  =  c,  ') 
dxy  =  e,S 


SIMULTANEOUS  QUADRATICS  579 

may  be  solved  by  the   methods  employed  for  solving  systems  of 
symmetric  equations. 

Ex.  5.    Solve  the  system  of  equations 

3x-{-2y=  11,     (1))  ,      ^.        ^ 

xy=    3.     (2)1^-     Given  System. 

Since  the  first  member  of  equation  (1)  is  the  binomial  sum  Sx  +  2y  we 
proceed  as  follows  to  obtain  an  equation  the  first  member  of  which  is  the 
binomial  difference  3x  —  2y. 

By  squaring  both  members  of  equation  (1)  we  obtain 

9  ar2+ 12  XT/ +  4 1/2=  121.  (3) 

Since  the  square  of  the  difference  3ic  —  2i/  differs  from  the  first  member 

of  equation  (3),  only  in  the  sign  of  its  "  middle  term  "  12  xy,  we  may,  by 

subtracting  from  the  members  of  equation  (3)  the  corresponding  members 

of  equation  (2),  each  multiplied  by  24,  obtain 

9a:«-12a;i/  +  4i/2  =  49.  (4) 

Equation  (4)  is  equivalent  to 

(3a;-2?/  +  7)(3a;-2i/-7)  =0.  •  (5) 

It  follows  that  the  given  system  of  equations  is  equivalent  to  the  follow- 
ing system  : 


3a; +  2^=11,     (0  )  tt      Equivalent 
.  2 1/  -  7)  =    0.     (5)  I   .  ■         Derived  ^ 


(3  a:  -  2  y  +  7) (3  a:  -  2 1/  -  7)  =    0.     (5)  j    .  Derived  System. 

The  solutions  of  this  system  of  equations  may  be  obtained  by  applying 
the  method  of  §  16. 

35.  The  solutions  of  a  system  of  two  symmetric  equations 
containing  two  unknowns,  x  and  ;/,  may  be  obtained  by  solving 
the  system  of  equations  obtained  by  substituting  particular  func- 
tions of  new  unknowns  for  the  given  set  of  unknowns,  x  and  ^, 
and  solving  the  resulting  equations  for  the  new  unknowns. 

It  will  sometimes  be  found  convenient  to  substitute  for  the 
given  unknowns,  x  and  i/,  the  sum  and  difference  respectively  of 
two  other  unknowns,  r  and  s,  that  is,  to  let  x  =  r  +  s  and 
y  =  r  —  s. 

Solving  the  resulting  system  of  equations,  we  obtain  values 
for  r  and  s  which,  when  substituted  in  the  assumed  equations 
x  =  r  -i-  s  and  y  =  r  —  s^  determine  the  values  of  the  unknowns, 
.'•  and  y. 


J  J    Equivalent 


580  FIRST  COURSE  IN  ALGEBRA 

Ex.  6.   Solve  the  system  of  symmetric  equations 

If  we  let     ~  '  [■  (3)  we   may  substitute  for  the  given  system   of 

y  —  r  —  s,  ) 

equations  (1)  an<l  (2)  the  equivalent  derived  system 

(r  +  .y+(r-s)*=l7,  (4)^ 

2r=    3,         (5) 

"  =  '•  +  ^'1(3) 

Eliminating  r  from  equations  (4)  and  (5),  we  obtain, 

16«*  + 2165'* -55  =  0.  (6) 

The  system  of  equations  consisting  of  equations  (6)  and  (5),  and  the 
assumed  equations  (3)  is  equivalent  to  the  given  system. 
The  solutions  of  equation  (6)  are  found  to  be 

«  =  ±  ^,        and        s  =  ±  i  \/—b5. 

Accordingly,  using  each  of  these  values  with  the  value  of  r  from  (5), 
we  have  the  following  pairs  of  values  for  r  and  s : 


Derived  System. 


::|:}o->::4:}("-)::i,^}(Ui.):.:j-^}ov.) 

Substituting  the  sets  of  values  (i.)  and  (ii.)  for  r  and  s  in  the  assumed 
equations  (3),  we  obtain  the  following  sets  of  real  values  of  x  and  y : 

^-M  and        ^=M 

^  =  1, )  2/  =  2.  j 

The  values  of  x  and  y  obtained  by  using  the  real  and  imaginary  values 
for  r  and  s  in  (iii.)  and  (iv.)  are  complex. 

All  of  the  values  thus  obtained  will  be  found  to  satisfy  the  given 
equations. 

Ex.  7.    Solve  the  system  of  equations 

,  I     /ox  \  I-    Given  Svstera. 

x  +  i/=    4.     (2)j 

,      Let  a:  =  r  +  s,         and         y  =  r  —  s.     (3) 

Using  the  equations  obtained  by  suV)stituting  r  +  s  for  x  and  r  —  s  for  y 

in  equations  (1)  and  (2)  with  the  assumed  equations  x  =  r  +  s  and  y  =  r  —  s, 

we  obtain  the  following  derived  system  of  equations  which  is  equivalent  to 

the  given  system  of  equations  : 


SIMULTANEOUS   QUADRATICS  581 

Equivalent 


3r^s  +  s^=    13,         (4) 
r=     2,         (5) 


x  =  r  +  s,\ 


'  I  (3) 

y  =  r  —  s.  )  ^  ^ 


Derived  System. 


From  equations  (4)  and  (5)  we  obtain,  eliminating  r,  and  transposing, 

s«+ 12  s -13  =  0.  (6) 

By  the  Factor  Theorem  we  find  that  s  —  1  is  one  of  the  factors  of  the 
first  member  of  equation  (6). 

Accordingly  equation  (6)  is  equivalent  to 

(s_l)(s2  +  5s  +  13)  =  0.  (7) 

The  solutions  of  (7)  are  found  to  be 

s  =  1,         and         s  =  ==-^ . 

Using  the  value  of  r  from  (5),.  we  obtain  the  following  pairs  of  values 
for  r  and  s  : 


r=2. 


("•)  .       ./-^hOii-) 


_i4..^/:r5T  r^-^        _i_^_5i 
'  = r "' 

Substituting  these  different  pairs  of  values  for  r  and  s  in  the  assumed 
equations  (3),  we  obtain  pairs  of  values  of  x  and  y  which  are  solutions 
of  the  given  system  of  equations. 

Exercise  XXV.     7.     Miscellaneous 
Solve- the  following  systems  of  equations: 

1.  X  =6-1/,  e.  x^  +  2x1/  +  f  =  36, 
f  =  —  20  —  x7/.               '  x^  —  2x//  +  f=16. 

2.  x'^dif=  16, 
X  +37/  =    8. 

3.  X  +  1/  =  x^j 
S(/-x  =  f. 

4.  x^  +  f=  178, 
X  —y  =    10. 

5.  (a;+v/)=  =  81, 
{x  -  yY  =9.  i^y  =  12. 


7. 

{x  -  1)0  +  1)  =  39, 
x  —  y=l2. 

8. 

«  —  3  ?/  =  5, 

icz/  =  4  a;^  —  2. 

9. 

!-'■ 

582  FIRST  COURSE  IN  ALGEBRA 

^^-  ^-^  =  1^'  21.  (x  +  1)0,  +  1)  =  54, 

1=    4,,  ic  +  v/  =  l3. 

^  22.  aj2  +  icy/  =      28, 

11.  02,  =  a^  xi/  +  f  =  ^  12. 

?  =  6^.  23.  a:-^-/=    7, 

•^  xi/  =  12. 

12.  ^±J^  =  5,                               24.  a!(a;  +  y)  -  20  =  0, 
^  y(y  +  aj)  -16  =  0. 


a^  =  64 


2o.     a;'^  +  i»y  =  42, 


13.  ^±J?  =  a,  K^-y)  =    5. 

^  ic~3,      a;4- v,       3  ' 

14.  ic  +  -  =  1,  «''  +  /        =  45. 

^  27.  ic*  +  xY  +  /  =  3, 

y  +  -  =  4.  a;2  -  i«y  +  7/2  =  1. 

J  J           28.  a:^  +      ^  +  -,,2  ^  133, 

15.  9a;+-  =  4=15ic x  +  ^xy  +  ;/  =    19. 

36ic_25y  29.  3ajy-a.y=       14, 

xy-\-x  —  y=^  122.  30.  aj^  ^- xy  =  20, 

11  a;    +    3,=    5. 

^^*                  ^~^"^'  31.  a^2  +  7/2  =  189  -  ^, 

1               1  5                     a;  +  ?/  =  9  +  Va^. 


ic  —  1       3/  —  1  3  *  32.  a;»    +  7/8    =:.  133, 

18    i4.i-^  «^^  +  «^/-    70. 

•  ar^  "^  7/2      49  '  33      ^3   +  ^8  ^  g5^ 

1       1__8  xy{x^y)^2^, 

^      y~~  l'  34.    aj«  —  7/3  =  26, 
19.  a;-^  —  7/-1  =  —  1,  x^y  —  aj/  =    6. 

""'  -  r'  =  -  5.  35.  x'^-.f^  56^ 


a; 


20.  7/(a;  ^y)=A 

x{x-y)=y.  ^      y- 


16 

a;^ 


SIMULTANEOUS   QUADRATICS  583 


y     ^ 


Ko  ^  +  y 

53.  — — ^  =  ax, 


y 

xy  =  h. 


X  x-{-  y  x''      0"- 

X  -\-  y  =\.  ^^  _  1 

37.  yis^-^y)^-x{y-^x)  —  ^xy,  ^y 

y{x  +  y)^x+y  =  2L  ^^    ax^hy^^ah, 

*  V«  +  V^      V'Xr-^/y      ^ 

x^  +  2/2  =  97^         52.  bx^-by  =  2, 

39.  a.- + //Va- =      40,  '  a6..y-l. 

a;-2  +  xy"-  =  1312. 

40.  (5i«-2/)(2/-'^^)  =  -l> 
{by  +  x){y  ->r  bx)  =  189. 

41.  .X.8  +  /  =  224, 

xy=    12.  .,        x  —  y     __ 

54.       -__     .--    8, 

42.  a;  4-  y  +  2Va^  +  ^  =    24,  'V/^J       VJ/ 

a;2  _j_  y2  ^  j3Q^  y^^  =  12. 

43.  a^'  +  r  -  '^('^  +  y)  =      ^'  «._,,_ 

,  +  ^  +  ..^  =  -7.     5s-:^^z^-^' 

44.  ^~  ,  =  r!'  Va^y  =  ^• 

56.  x^-=Q>x+  4.y, 

y^  =  4.X+  6^. 

57.  x"^  =  cx  +  dy, 
y'^  z=  dx  -{■  cy. 

58.  a^  +  /  =  33. 
a;  +  2/   =    ^* 

59.  ^^  +  /  =  97, 

a-    +  7/    =     5. 

60.  x'-\-y^  =  97, 
X  -y   '^     1. 


7/-l_ 

a;-  1 

/  +  .y  +  1  _ 

^3' 
43 

a;^  +  a;  +  1 

21 

45. 

{x-y){x'-y'^^-^'^. 

46. 

ax  =  by, 

^^  +  /  =  c. 

47. 

VX  =  gy, 
{p-\-q)x-{p-'q)y  =  r. 

48. 

a(:x  +  y)  =  b(x-y)=xy. 

49. 

x"  —  y"-  =  «, 

a;4  _  7/*  =  6. 

584  FIRST  COURSE  IN   ALGEBRA 


1    _1  =_L-,  62    ^'  +  ^  +  ^-.^  =  111 

X       y       X  —  y'  '  y^      'it?      y      X       m 

J__J._J_  1 


Simultaneous  Kquations  Involvingr  Decimal  Fractions 

Solve  the  following  systems  of  equations  and  find  approximate 
values  of  the  unknowns  which  are  correct  to  three  places  of  decimals: 

63.  a;  +  ?/  =  2.8,  72.  o?  +  /  =  9.0625, 

xy  =  1.87.  X  -\-y  =  3.25. 

64.  a;  +  y  =  .051,  73.  4  a^  -  /  =  2.03, 

xy  =  .000518.  x-\-y=.% 

65.  10a;  —  10^  =  .5,  74.  .2a!^  -  xy  =  —  .742, 

10a;^  =  .126.  x+.ly=      .82. 

66.  .5a;  — .17=    .1?/,  75.  2.5a;  +  .3?/ =  7, 

\Oxy=  1.2.  .5a;y  =  6. 

67.  .001  xy=  1.075,  76.  .1  x  +  y  =  8, 
,\x  —  .\y=  1.8.  xy  =  2A, 

68.  a?-\-f  =  .89,  77.  a;^  +  10/  =  14.49, 

10a:ry  =  4.  x  ■\- y  =    1.5. 

69.  100  ar^  +  100  r  =  65,  78.  5  a;^  +  ^2  ^  9.2, 

a;;/  =  .28.  xy  =    .6. 

70.  .01  a;  +  .01  y  =  .0015,  79.  a;»  -  /  =  .056. 

.lar^H- .ly'-'  =  . 00125.  x  —  y  =  .2 

71.  ar^  +  /  =  11.3,  80.  3  a;-'  —  /  =  299.99. 

a;  -J-  7/   =r      4.4.  xy  =     1. 

Systems  of  Three  or  More  Equations  Containing 
Three  or  More  Unknowns 

36.  The  solution  of  a  system  of  three  simultaneous  equations 
containing  three  unknowns  can  be  made  to  depend  upon  the  solution 
of  a  quadratic  equation  only  in  exceptional  cases. 

There  are  certain  systems  of  special  equations  the  solutions  of 


X 

-     y-2z  = 

0, 

X 

-^2y-h^::  = 

11, 

X 

2  +    2/2  4.    .2  ^ 

21. 

SIMULTANEOUS   QUADRATICS  585 

which  can  be  made  to  depend  upon  the  solutions  of  quadratic 
equations. 

37:  If  a  system  of  three  or  more  simultaneous  equations  contains 
one  and  only  one  equation  of  the  second  degree  with  reference  to 
the  unknowns,  the  remaining  equations  being  all  of  the  first  degree, 
the  solution  of  the  system  can  be  made  to  depend  upon  the  solution 
of  a  quadratic  equation  containing  one  unknown. 

Ex.  1.   Solve  the  system  of  equations    . 

(1)] 

(2)  V  I.  Given  System. 

(3)) 

The  unknowns  x  and  y  may  be  expressed  in  terms  of  the  remaining 
unknown,  z,  as  follows  : 

Employing  equations  (1)  and  (2),  and  eliminating  y,  we  obtain 

3a; -2  =11.  (4) 

Hence,  x  =  — (5) 

o 

From  equations  (1)  and  (2),  eliminating  x,  b}'^  subtraction,  we  obtain 

3y  +  5z=U.  (6) 

Hence,  y  =  — ^ .  (7) 

Substituting  these  "  expressed  "  values  for  x  and  y  in  equation  (3),  we 
obtain  a  quadratic  equation  containing  z  alone,  from  which  the  values  of  z 
are  found  to  be  1  and  |-|. 

Substituting  these  values  for  z  in  equations  (5)  and  (7),  we  obtain  cor- 
respondinf:^  values  of  a:  and  y. 

Accordingly  the  solutions  of  the  given  system  are  found  to  be 

x  =  4A  ^  =  W' 

y  =  2A         and        i/  =  f , 

^=lj  ^-M- 

These  sets  of  values  will  be  found  to  satisfy  the  given  equations. 
Ex.  2.    Solve  the  system  of  equations 

xy  =  2,i\)\ 

yz  =  4,  (2)  yi.    Given  System. 

zx  =  ii.  (3)  J 

We  may  obtain  the  following  equation,  the  members  of  which  are  tbo 


586  FIRST  COURSE   IN  ALGEBRA 

continued    products    of   the    corresponding   members   of  the   three   given 
equations : 

xhjh^  =  64.  (4) 

From  equation  (4)  we  obtain 

ryz  =  ±S.  (5) 

This  result  may  be  interpreted  as  representing  the  set  of  two  equations, 
xyz  =  +  8  and  xyz  =  —  8,  which,  taken  together,  are  equivalent  to  equa- 
tion (4). . 

Dividing  the  members  of  (5)  by  the  corresponding  members  of  equations 
(1),  (2),  and  (3)  respectively,  we  obtain  the  following  results : 
2  =  ±4,        x  =  ±2,        y  =  ±l. 

Since  the  different  members  of  all  of  the  given  equations  are  positive, 
and  the  first  members  contain  two  factors  each,  it  follows  that  the  signs 
of  these  factors  must  be  like.  Accordingly,  we  may  arrange  these  values 
in  sets,  as  follows : 

X  = 


=  1,  [-        and        2/  =  -  1,1 
=  4,J  z  =  -4.} 


These   sets  of  values  are  found  by  substitution  to  satisfy  the  given 
equations. 

Ex.  3.   Solve  the  system  of  equations 

(x  +  i,)(x  +  .)=    4,     (1)^ 
.  (2/  +  2)(y  +  x)  =  16,     (2)  U.    Given  System. 

(;2  +  x)(s  +  i/)  =  36.     (3)  J 

Multiplying  together  the  corresponding  members  of  the  given  equations 
and  taking  the  square  roots  of  the  results,  we  obtain 

(x  +  yXy  +  2)(2  +  a:)  =  ±  48.  .(4) 

Using  the  members  of  the  given  equations  as  divisors  with  the  corre- 
sponding members  of  (4),  we  obtain 

y-{.z  =  ±  12,     (5)         x  +  y  =  ±3,     (6)         ^  +  x  =  ±  f ,  (7) 

from  which  the  values  of  x,  y,  and  z  may  be  obtained. 
Ex.  4.     Solve  the  system  of  equations 
x^-\-2yz=l,     (1)^ 
y^  +  2zx=l,     (2)  VI,     Given  System. 
z''  +  2xy=2.     (3)  J 

Adding  the  corresponding  members  of  the  given  equations,  we  obtain, 

r^  +  y^-\-^^^  -h^y^^  +  ^zx  -\-2xy  =  4.  (4) 

Hence,  [x  +  y  +  z  +  2]lx -^  y  +  z  -  2]  =  0.  (5) 


SIMULTANEOUS   QUADRATICS  687 

Subtracting  the  members  of  equation  (2)  from  the  corresponding  mem- 
bers of  (1),  we  obtain 

x^-y^-\-2yz-2zx  =  0.  (6) 

Or,  '  (x-y)[x-\-y-2z-]  =  0.  (7) 

The  system  composed  of  equations  (5)  and  (7),  and  any  one  of  the  given 
equations,  such  as  (1),  is  equivalent  to  the  given  system  of  equations. 

[,  +  ,  +  .  + 21[.  +  ,  +  .- 2]  =  0      (5)1  ^.^^^^^^ 

(x-,)[x  +  ,-2.]  =  0,  II.       'derived  System. 

x^  4-  2yz  =  1.     (l)j  "^ 

Equating  the  factors  of  the  first  members  of  equations  (5)  and  (7)  sep- 
arately to  zero,  and  applying  the  method  of  §  16,  we  may  separate  the 
derived  System  II.  into  a  group  of  four  derived  systems,  which  taken  to- 
gether are  equivalent  to  System  II. 

Solving  these  systems  separately,  the  solutions  of  the  given  system  of 
equations  may  be  obtained. 

Ex.  5.    Solve  the  system  of  equations 
x  +  xy  +  y=z    3,     (l)\ 
y  +  2/2  +  2  =    8,     (2)  M .    Given  System. 
x-^xz+z=15.     (3)J 
From  the  first  equation  we  may  obtain  the  value  of  x,  expressed  in  terms 
of  y,  as  follows : 

xO-  +  y)  +y  =  3. 

Hence,  x  = •  (4) 

1+3/ 

Substituting  this  value  for  x  in  equation  (3),  we  obtain  an  equation  con- 
taining y  and  z  the  members  of  which  may  be  combined  with  those  of  equa- 
tion (2)  to  obtain  the  values  of  y  and  z. 

The  values  of  x  may  be  obtained  by  substituting  in  equation  (4). 
Ex.  6.   Solve  the  system  of  equations 
x^-yz  =  a,     (1)^ 

y^-zx  =  b,     (2)  V  I.  Given  System. 
z^-xy  =  c,     (3)j 
If,  from  the  squares  of  the  members  of  equation  (1)  we  subtract  the 
product  of  the  corresponding  members  of  equations  (2)  and  (3),  we  shall 
obtain  equation  (4). 

Equations  (5)  and  (6)  may  be  obtained  in  a  similar  way. 

x[x^ -^  yi  ^  z^  -  3 xyz]  =  a^  -  be.  (4) 

y  [a;8  4.  ys  +  ?j8  _  3  xyz]  =  &2  -  ca.  (5) 

2  [a;3  +  y8  +  z^  -  3  xu;i\  =  c^  -  ah.  (6) 


588  FIRST  COURSE  IN  ALGEBRA 

From  equations  (4)  and  (6)  we  obtain  by  division  the  value  of  the  fraction 
xjz.  Similarly,  from  equations  (5)  and  (6)  we  obtain  the  value  of  the 
fraction  yf  z. 

From  the  equations  thus  obtained  we  may  find  the  expressed  values  of 
X  and  y  in  terms  of  z. 

Substituting  for  x  and  y  in  equation  (1)  their  expressed  values  thus  found, 
we  obtain  the  value  of  z  in  terms  of  the  known  numbers  a,  6,  and  c,  in  the 
form  of  a  fraction  having  an  irrational  denominator, 

±  (c^  -  ah) 
^a8  +  6«H-c8-3a6c  ^^ 

Either  by  substituting  this  value  for  z  in  the  remaining  equations,  which 
may  then  be  solved  for  x  and  y,  or  by  repeating  the  process  above  with  dif- 
ferent pairs  of  equations,  we  obtain  expressions  of  the  same  type  as  (7)  for 
X  and  1^. 

Exercise  XXV.     8 

Find  sets  of  values  which  satisfy  each  of  the  following  sys- 
tems of  equations: 


I.  xy=       30, 

7. 

yz  =  6c, 

10. 

xy^z^  =  -  24, 

yz  =  -  60, 

^  .  3^ 

yz^             4 

xz  =  —  50. 

«+!='• 

x=    r 

2.yz  =  a\ 

X        z         ^ 

x'y  _        9^ 

xz  =  b'. 

-  +  -  =  1. 

a      c 

z              2 

xy  —  f^. 

xy     _    4.^ 

3.  ^z=    2, 

11. 

x  +  y        3* 

fz=    1. 

8. 

x'-^y^^  13, 

z^x  =  32. 

ar^  +  ;:«  =  34, 

yz     _  12  ^ 
y  +  z       5  ' 

4.  xh)z  =  a, 

y''  +  z^  =  29. 

xys^  =  c. 

zx           3 

y  +  z  =  ~^ 

X 

2;  +  a;         2 

5.  yz  =  '2y-\-  4tz, 

9. 

12. 

"^"    =2, 

ZX=     4:Z     +     Xy 

jc  +  y 

xy  =  x+  2y. 

^  +  ^  ^  7/ 

JK?/;:;        3 

6.  x(:y  +  z)  =  5, 

y 

x  +  z      2 

y(x  +  z)  =  8, 
z{x  +  y)  =  ^. 

x^-y=-' 

a;?/2;    _6^ 
y  +  z     5 

SIMULTANEOUS  QUADEATICS  589 

13      ^.y-    =  a  ^^'  (^  +  "^^^^  +  ^)  =  ^'' 

xyz    _  0  +  ^)(a;  +  z)  =^  c^. 

y  +  z        '  23.  a;  +  i/  +  2;  =  a, 

^yz    ^  ^^  a;^  =  ^, 

a;  +  2;        *  xyz  =  c. 

^.+    .        5  2i.x  +  y    =    3, 

y-±jL^^,  x^+s  =  8. 

xyz         12 

^.  +  ^.      17  25.  «    +y   +    ^=      9, 

=  T^*  ajv  +  V2;  +  ;:^£c  —     26, 

"^^    ^'  j+;._,.  =^3. 

15.  («  +  \){y  +  1)  =  8,  26.  a;^  -  (2,  -  ;^)2  =  1, 
(^  +  !)(;,  +  1)  =  24,  /  -  (;2^  -  a^)'  =  4, 
(;^+l)(a;+  1)  =  12.  z''-{x-yy  =  ^. 

16.  a;(2  -  ?/)  =  16,  27.  a;(?/  +  z)  +  3  =  0, 
3/(2- Is)  =  9,  y{z-^x)-¥21  =  0, 
z(2  —  x)=    4.  2^(3  X—  y)=^0. 

17.  ar^  =  ?/2;,  28.  x  +  xy  +  y=15, 
x  +  y  +  z=    21,  y  +  yz  +  z  =  24., 

xijz  =  2ie.  x  +  xz  +  z  =  S5. 

18.  xy  +  xz  +  yz  =  3,  29.  x^  —  yz  =  2, 

x  —  y  =  2y  y"^  —  zx  =  4., 

y  —  z=\.  z^  —  xy=l. 

19.  a;(a;  +  ?/  +  5r)  ==  6.  30.  g?  -yz  =  49, 
y(a;  +  y  +  4  =  12»  y  — 2:a;=  1, 
2^(a;  +  y  +  4  =  18.  z^  —  xy=  79. 

20.  (3/  +  z)(x  +  7/  +  2;)  =  6,  31.  7/2;  +  a;  +  y  =  -  9, 
(z  +  x)(x  +  y -\-z)=  8,  a;2;  +  2/  =  -5, 
(a;  +  2/)(a;  +  ^  +  ^)  =  — 6.  a^^  =      2. 

21.  a:^^  =  c(a;  +  y  +  2;),  32.  x" -{■  xy  +  y'' -=  3, 
3^;^  =  flr(a;  +  3/  +  2;),  /  4-  ^^  +  z^  =  7, 
a!;r  =  ^'(a;  +  7/  +  ;s).                          2;^  +  ^^a;  +  a;^  =  7. 


590  FIRST  COURSE  IN  ALGEBRA 

X      y      z        3 

xyz  =  1. 

Exercise  XXV.    9 

Solve  the  following  problems  employing  conditional  equations 
containing  two  or  more  unknown  quantities : 

1 .  Find  two  numbers  the  sum  of  which  is  40,  and  the  product  of  which 
is  256. 

2.  Find  two  numbers  the  difference  of  which  is  15,  and  the  sum  of  the 
squares  of  which  is  293. 

3.  Find  two  numbers  the  difference  of  which  is  6,  and  the  product  of 
which  is  247. 

4.  Find  two  numbers  the  sum  of  which  is  40,  and  the  product  of  which 
is  391. 

5.  Find  two  numbers  the  prwluct  of  which  is  96,  and  the  sum  of  the 
squares  of  which  is  208. 

6.  Find  two  numbers  the  difference  of  the  squares  of  which  is  112,  and 
the  square  of  the  difference  of  which  is  64. 

7.  Find  two  numbers  the  sum  of  which  is  10,  and  the  sum  of  the  cubes 
of  which  is  370. 

8.  The  product  of  two  numbers  is  64,  and  the  quotient  obtained  by 
dividing  the  gre.ater  number  by  the  less  is  4.     Find  the  numbers. 

9.  If  the  sum  of  two  numbers  is  divided  by  the  less  number,  the  quo- 
tient is  4;  the  protluct  of  the  numbers  is  27.     Find  the  numbers. 

10.  Find  two  numbers  the  sum  of  which  is  20,  such  that  the  snm  of 
the  quotients  obtained  by  dividing  each  number  by  the  other  is  17/4. 

11.  Find  two  numbers  such  that,  if  each  be  increased  by  1,  the  product 
is  124,  and  the  product  obtained  by  multiplying  the  first  number  by  a 
number  less  by  one  than  the  second  number  is  60. 

12.  Find  two  numbejps  the  sum  of  which  is  twice  their  difference,  and 
the  difference  of  the  squares  of  which  is  200. 

13.  The  product  of  two  numbers  is  12,  and  the  sum  of  their  squares  is 
five  times  the  sum  of  the  numbers.     Find  the  numbers. 

14.  Find  two  numbers,  of  which  the  sum  is  14,  which  are  such  that  the 
product  of  the  first  and  the  reciprocal  of  the  second,  increased  by  the 
product  of  the  second  and  the  reciprocal  of  the  first,  is  25  /  12. 


PROBLEMS  591 

15.  The  sum  of  two  numbers  is  30,  and  the  sum  of  the  quotients  result- 
ing from  dividing  each  number  by  the  other  is  82/9.     Find  the  numbers. 

16.  The  first  of  two  numbers  is  ten  times  the  reciprocal  of  the  second, 
and  the  sum  of  the  second  number  and  ten  times  the  reciprocal  of  the  first 
is  equal  to  the  square  of  the  second  number.     Find  the  numbers. 

17.  Find  two  fractions  such  that  the  sum  of  the  first  fraction  and  the 
reciprocal  of  the  second  is  equal  to  2,  and  the  sum  of  the  second  fraction 
and  the  reciprocal  of  the  first  is  8/3. 

18.  Find  two  numbers  the  sum  of  which  is  36,  and  half  the  product 
of  which  is  equal  to  the  cube  of  the  less  number. 

19.  Find  a  fraction  the  value  of  which  is  3/4,  and  the  product  of  the 
numerator  and  denominator  of  which  is  48. 

20.  If  a  certain  two-figure  number,  the  sum  of  the  figures  of  which 
is  12,  be  multiplied  by  the  units'  figure,  the  product  is  375.  What  is  the 
number  1 

21.  A  number  expressed  by  two  figures  is  equal  to  four  times  the  sum  of 
the  figures.  The  number  formed  by  writing  the  figures  in  reversed  order 
exceeds  three  times  the  product  of  the  figures  by  the  square  of  the  figure  in 
tens'  place  of  the  given  number.     Find  the  number. 

22.  If  it  requires  240  rods  of  fence  to  enclose  a  rectangular  field  of  20 
acres,  what  are  the  dimensions  of  the  field  1 

23.  A  rectangular  field  contains  30  acres.  By  increasing  its  length  by 
40  rods  and  diminishing  its  width  by  4  rods,  the  area  is  increased  by  6 
acres.     What  are  its  dimensions  ? 

24.  The  length  of  the  fence  around  a  rectangular  field  is  274  yards,  and 
the  distance  measured  diagonally  from  corner  to  corner  is  97  yards.  What 
is  the  area? 

25.  Thirty-two  yards  of  the  fence  about  a  rectangular  field  which  is  184 
yards  long  and  76  yards  wide  are  destroyed.  What  must  be  the  dinien- 
.^ions  of  a  rectangular  field  in  order  that  the  length  of  fence  remaining  shall 
enclose  the  same  area  as  before  ? 

26.  A  property  owner  wishes  to  use  the  material  from  a  stone  wall 
enclosing  a  field,  which  has  the  form  of  a  rectangle  80  rods  long  and  60  rods 
wide,  to  build  another  wall  greater  by  16  rods  which  shall  enclose  a  second 
liact  of  land  which  has  the  form  of  a  rectangle  having  the  same  area  as  the 
first.     Find  the  dimensions  of  the  second  tract  of  land. 

27.  In  widening  a  street,  a  strip  of  land  6  feet  in  width  was  removed 
from  the  entire  frontage  of  a  tract  containing  28,800  square  feet.  By 
increasing  the  frontage  of  the  reduced  lot  by  8  feet,  the  entire  area  became 
the  same  as  before.     Find  the  original  dimensions  of  the  land. 

28.  It  is  observed  that,  if  a  guy  rope  which  is  attached  to  a  stake  7  feet 


692  FIRST  COURSE   IN  ALGEBRA 

from  the  foot  of  a  derrick  were  lengthened  by  15  feet,  it  would  rea€h  to  a 
stake  32  feet  from  the  foot  of  the  derrick.  Find  the  height  of  the  derrick 
and  the  length  of  the  rope. 

29.  A  tract  of  10  acres  of  land  is  enclosed  by  a  certain  length  of  fence  in 
such  a  way  that  there  are  two  separate  lots,  each  in  the  form  of  a  square, 
so  situated  that  the  side  of  the  smaller  lot  forms  a  part  of  the  side  of  the 
larger  lot.  It  is  observed  that  the  fence  may  be  rebuilt  to  enclose  a  single 
lot  in  the  form  of  a  square  containing  5|  acres  more  than  the  original  lots. 
Find  the  dimensions  of  the  original  lots. 

30.  At  an  entertainment  $750  was  realized  from  the  sale  of  seats.  For 
each  reserved  seat  twenty-five  cents  more  was  charged  than  for  an  un- 
reserved seat,  but  the  sale  of  the  unreserved  seats  yielded  the  same  total 
amount  as  that  of  the  reserved  seats.  Find  the  total  number  of  seats  sold, 
if  the  number  of  unreserved  seats  exceeded  the  number  of  reserved  seats 
by  125. 

31.  Three  stone  crushers  working  together  can  crush  a  certain  amount 
of  stone  in  a  week.  The  first  machine  has  a  capacity  twice  as  great  as  that 
of  the  second,  but  working  alone  would  require  one  week  more  than  the 
third  machine  to  perform  the  work.  What  time  would  each  require, 
working  alone  ? 

32.  The  sum  of  a  fraction  and  its  reciprocal  is  equSil  to  the  numerator 
increased  by  the  reciprocal  of  twice  the  denominator ;  and  the  difference 
between  the  reciprocal  of  the  numerator  and  the  reciprocal  of  the  de- 
nominator is  equal  to  the  reciprocal  of  twice  the  denominator.  What  is 
the  fraction  ? 

33.  Find  a  fraction  such  that,  if  its  numerator  be  increased  by  3  and  its 
denominator  diminished  by  3,  the  result  is  the  reciprocal  of  the  fraction  ; 
but  if  the  denominator  be  increased  by  3  and  the  numerator  diminished  by 
3,  the  result  will  be  1^  less  than  the  reciprocal  of  the  fraction. 

34.  Find  a  fraction,  the  value  of  which  is  2/3,  such  that  if  the  numerator 
be  diminished  by  the  reciprocal  of  the  denominator  and  the  denominator 
be  increased  by  the  reciprocal  of  the  numerator,  the  value  of  the  fraction 
will  be  multiplied  by  149/151. 

35.  Find  two  numbei-s  the  sum  of  the  cubes  of  which  is  133,  and  the  sum 
of  the  squares  of  which  diminished  by  their  product  is  19. 

36.  Find  two  numbers  of  which  the  sum  multiplied  by  the  product  is 
equal  to  30,  and  the  sum  of  the  cubes  of  which  is  35. 

37.  A  number  is  expressed  by  three  figures,  the  sum  of  which  is  10. 
The  middle  figure  exceeds  the  sum  of  the  other  two  by  2,  and  the  sum  of 
the  squares  of  the  separate  figures  is  seven  times  the  figure  in  tens'  place, 
increased  by  the  sum  of  the  other  two.     Find  the  number. 


PROBLEMS  593 

38.  Two  boats  leave  simultaneously  the  opposite  shores  of  a  river  which 
is  2}  miles  wide,  and  pass  each  other  in  15  minutes.  The  faster  boat  com- 
pletes the  trip  6|  minutes  before  the  other  reaches  the  opposite  shore. 
Find  the  rates  of  the  boats  in  miles  per  hour. 

39.  A  train  starts  from  a  certain  station  to  make  a  trip  of  180  miles, 
travelling  uniformly.  Forty -five  minutes  later  a  faster  train,  also  travelling 
uniformly,  starts  from  the  same  station,  and  after  travelling  two  hours  and 
fifteen  minutes  reaches  the  station  which  the  first  train  had  passed  thirty-six 
minutes  previously.  The  speed  of  the  second  train  is  now  increased  by 
four  miles  an  hour,  with  the  result  that  the  trains  reach  the  terminus  at  the 
same  time.     Find  the  rates  in  miles  per  hour,  at  which  they  started. 

40.  Fifteen  hours  after  an  ocean  steamship  leaves  the  American  shore  to 
make  a  voyage  of  3300  miles  at  a  certain  average  uniform  rate,  a  second 
ship  starts  for  America  from  the  opposite  shore.  The  two  ships  meet  after 
the  second  ship  has  been  out  7 If  hours,  and  they  complete  their  trips  at  the 
same  instant.     Find  the  rates  of  the  ships  in  miles  per  hour. 

41.  An  express  train,  an  electric  car,  and  an  automobile  all  leave  a  given 
place  for  a  certain  destination.  The  express  train  travels  9  miles  an  hour 
faster  than  the  electric  car,  and  the  automobile  13  miles  an  hour  faster  than 
the  express.  The  express  starts  one  hour  after  the  electric  and  39  minutes 
before  the  automobile.  If  all  arrive  at  the  end  of  their  journeys  at  the  same 
time,  find  the  distance  and  the  rates  of  travelling  in  miles  per  hour. 

42.  Sighting  an  enemy's  war  vessel  at  a  distance  of  10  miles,  a  submarine 
boat  starts  toward  it,  running  on  the  surface  at  a  certain  uniform  rate  which 
exceeds  its  speed  when  submerged  by  6  miles  an  hour.  At  a  certain  point 
in  its  course  it  dives  beneath  the  surfiice,  and  when  submerged  at  a  distance 
of  one-half  mile  from  the  battleship,  a  torpedo  is  discharged  which  travels 
at  the  rate  of  a  mile  in  two  minutes.  It  is  observed  from  the  shore  that  the 
time,  measured  from  the  instant  the  submarine  starts  until  the  explosion 
takes  place,  is  exactly  44|  minutes.  Returning  immediately  after  deliver- 
ing its  torpedo,  and  travelling  the  entire  distance  under  water,  the  time 
required  is  one  hour,  three  and  one-third  minutes.  Find  the  rate  of  the 
submarine  on  the  surface  in  miles  per  hour  and  also  the  distance  from  the 
starting-point  of  the  spot  at  which  it  sank  beneath  the  surface. 

43.  A  torpedo  boat,  on  being  discovered  1^  miles  from  port,  immedi- 
ately turns  and  tries  to  escape.  One  minute  later  a  torpedo-boat  destro3'er 
is  sent  out  and  this  overtakes  it  after  a  run  of  9^  miles.  If  the  rates  of  the 
torpedo  boat  and  destroyer  could  have  been  increased  by  6  miles  an  hour 
and  2  miles  an  hour  respectively,  when  the  distance  between  the  two  boats 
was  reduced  to  1  mile  the  torpedo  boat  would  have  escaped  to  its  squadron 
I  <  ».V  miles  away.     Find  the  rates  of  the  boats  in  miles  per  hour. 

38 


594  FIRST  COURSE  IN  ALGEBRA 


CHAPTER  XXVI 

RATIO,  PROPORTION,  AND  VARIATION 

I.     Ratio 

1.  The  ratio  of  one  number  to  another  of  the  same  kind  is  the 
quotient  obtained  by  dividing  the  first  number  by  the  second. 

The  ratio  of  «  to  ^  may  be  expressed  by  any  symbol  of  division. 

E.  g.    a-^b;  J- ;  a/b ;  or  by  a  :  6. 
The  ratio  of  6  to  3  is  |  or  2. 

2.  In  a  ratio  a  :  b  (read,  "the  ratio  of  a  to  ^"),  a  is  called  the 
first  term  or  antecedent  of  the  ratio,  and  b  the  second  term 
or  consequent. 

3.  Since  a  ratio  has  been  defined  as  a  fraction,  it  follows  that  all 
of  the  properties  of  fractions  apply  also  to  ratios. 

„  am  _a  1 2  _  3 

■  °*  bm~V         l6~4" 

4.  According  as  a  >  h  or  a  <  b  the  ratio  a  :  b  is  said  to  be  a 
ratio  of  greater  inequality,  or  a  ratio  of  less  inequality. 

5.  The  values  of  two  ratios  may  be  compared  by  expressing  them 
as  fractions  and  then  reducing  the  iractions  to  equivalent  fractions 
having  a  common  denominator. 

E.  g.    Compare  the  values  of  the  ratios  2  :  3  and  7  :  8. 

We  have         2:3  =  f  =  :^f;         also         7  :  8  =  |  =  f  i . 

Hence  since  |^  >  ^|,  it  appears  that  7  :  8  >  2  :  3. 

6.  Two  or  more  ratios  are  said  to  be  compounded  if  their 
corresponding  terms  are  multiplied  together. 

E.  g.    acx  :  bdy  is  compounded  of  a  :  b,  c  :  d,  and  x  :  y. 


RATIO  595 

7.  A  duplicate  ratio  is  the  ratio  formed  by  compounding  two 
equal  ratios. 

E.  g.    a^  :  &2  is  the  duplicate  ratio  of  the  ratio  a  :  h. 

8.  A  triplicate  ratio  is  the  ratio  formed  by  compounding  three 
equal  ratios. 

E.  g.    a^  :  b^  is  the  triplicate  ratio  of  the  ratio  a  :  h. 

9.  The  inverse  ratio  of  a  ratio  is  obtained  by  interchanging  the 
antecedent  and  the  conseciuent. 

E.  g.    a  :  h  and  b  :  a  are  inverse  ratios. 

10.  Principle :  A  ratio  of  greater  inequality  is  diminished,  and 
a  ratio  of  less  ineqiialiti)  is  increased,  by  the  addition  of  the  same 
positive  number  to  each  of  its  terms. 

If  a,  b,  and  x  are  any  positive  numbers,  the  ratio  j  is  greater  than 

or  less  than  the  ratio  -. according  as  a  is  greater  than  or  less 

than  b. 

Since  b  and  x  are  both  positive,  the  denominator  b{b  +  x)  is 

positive,  and  the  value  of  the  second  member  of  (1)  is  positive  or 

negative  according  as  the  factor  {a  —  b)  of  the  numerator  is  positive 

or  negative,  —  that  is,  according  as  a  is  greater  than  or  less  than  b. 

CL  a  ~l~  X 

The  value  of  7  is  greater  than  or  less  than  the  value  of  t— — 

b  b  +  X 

according  as  the  value  of  the  difference  a  —  b  is  positive  or  negative. 

In  a  similar  manner  it  may  be  shown  that  a  ratio  of  greater 

inequality  is  increased,  and  a  ratio  of  less  inequality  is  diminished,  by 

subtracting  the  same  positive  number  from  each  of  its  terms. 

11.  If  two  concrete  quantities  of  the  same  kind  can  each  be  ex- 
pressed in  whole  numbers  in  terms  of  some  unit  of  measure,  this 
common  unit  is  called  a  common  measure  of  the  two  given  quan- 
tities, and  the  two  given  quantities  are  said  to  be  commensurable. 
(See  Chap.  XVIIL  §  8,  and  .Chap.  XX.  §  1.) 

12.  The  number  which  expresses  the  number  of  times  a  given 


596  FIRST  COURSE  IN  ALGEBRA 

unit  is  contained  in  a  given  quantity  of  the  same  kind  is  called  the 
iiiiinerical  measure  of  the  quantity  with  respect  to  the  given 
unit. 

13.  The  ratio  of  two  concrete  quantities  of  the  same  kind, 
which  can  both  be  expressed  in  terms  of  the  same  unit  by  means  of 
two  rational  numbers,  is  defined  to  be  the  ratio  of  their  numerical 
measures. 

E.  g.    The  ratio  of  2J^  feet  to  3^  feet  is  the  ratio  -y,  or  ||.     By  einploy- 

ini^  A  ^^  ^  ^^^^  ^'^  *^  common  unit  of  measure,  we  can  express  2^  feet  and 
3^  feet  in  terms  of  this  common  unit  by  the  numbers  35  and  48  respectively. 

14.  Two  quantities  of  the  same  kind  are  said  to  be  incoinnien- 
surable  if  both  cannot  be  expressed  in  whole  numbers  in  terms  of 
a  common  unit  of  measure. 

The  exact  value  of  the  ratio  of  two  incommensurable  quantities 
cannot  be  expressed  in  terms  of  a  whole  number,  or  of  a  fraction 
the  numerator  and  denominator  of  which  contain  a  finite  number 
of  figures. 

An  approximate  value  mat;  he  found  which  will  difier  from  the 
true  value  of  the  ratio  of  two  incommensurable  quantities  by  less  than 
any  assignable  value^  however  small. 

(The  foUowing  proof  may  be  omitted  when  the  chapter  ia  read  for  the  first  time.) 

Let  A  and  B  i-epresent  two  incommensurable  quantities  of  the  same 
kind.  It  is  always  possible  to  find  two  whole  numbers  of  which  the  ratio 
differs  from  the  true  value  of  the  ratio  of  A  to  B  by  as  small  a  value  as  we 
please. 

For,  if  we  separate  the  lesser  of  the  two  quantfties,  say  B,  into  any  integral 
number  n  of  equal  parts,  then,  since  A  and  B  are  assumed  to  be  incom- 
mensnrable,  it  will  follow  that  when  A  is  divided  by  this  unit  B/n,  there 
will  l)e  a  remainder  which  is  less  than  one  of  these  nth  parts  of  B. 

Suppose  that  one  of  these  nth  parts  of  B  is  contained  in  A  more  than 
m  times  and  less  than  m  +  1  times. 

Then  the  true  value  of  A  jB  will  lie  between  the  approximate  values 

m       -m+1       ,,.     m      A       m+1 

—  and  5   that  is,  —  <  >,  < " " 

n  n  n       b  n 

It  follows  that  either  approximate  value,  say  m/w,  differs  from  the  true 
value  of  AjB  by  a  value  less  than  that  by  which  it  differs  from  the  other 
approximate  value,  say  (rn  +  ^)Jn. 


RATIO  597 

The  difference  1/n,  between  the  approximate  values  mjn  and  (m+  l)/n, 
can  be  made  as  small  as  we  please  by  taking  n  great  enough,  but  it  can 
never  he  made  .equal  to  zero. 

Accordingly,  an  approximate  value  may  be  found  which  will  differ  by 
less  than  any  assignable  value  from  the  true  value  of  the  ratio  of  two  given 
incommensurable  quantities,  A  and  B. 

It  should  be  observed  that  the  commensurable  ratios  min  and 
{m  +  \)l?ij  which  approximate  to  the  true  value  of  the  ratio  of  the 
incommensurable  numbers  A  and  B,  define  an  incommensurable 
number. 

Accordingly,  the  fixed  valm  which  is  the  ratio  of  two  incommen- 
surable quantities  is  called  an  incommensurable  ratio. 

E.  g.  The  numbers  y^  and  3  are  incommensurable  with  respect  to  each 
other  ;  hence  their  ratio  — ^  is  an  incommensurable  ratio. 

The  incommensurable  numbers  2-y/3  and  5/y/3  are  commensurable  with 

2a/3 
respect  to  each  otJier,  since  their  ratio  —~=  is  equal  to  f  • 

o-Y^  3 

It  may  be  shown  by  applying  the  Principles  of  Variables  and 
Limits  that  two  incommensurable  ratios  are  equal  if  their  approxi- 
mate values  remain  equal  as  the  unit  of  measure  is  indefinitely 
diminished. 

Exercise  XXVL     1 

Write  each  of  the  following  ratios  in  simplest  form  : 

1.  10  :  12.  4.    x'-.xij.  7.    i  :  i  •  10. 

2.  25  :  30.  5.   abc -.  bed.  8.    ^  :  i-  11. 

3.  6«:8a.  6.    la^-.'i^ab.  9.    |  :  f-  12.    -g 

Find  the  ratio  compounded  of 

13.    4:5  and  10  :  6.       14.  2:3  and  4  :  9.       15.  21  :  4  and  10  :  7. 

16.  6:7  and  the  duplicate  of  2  :  3. 

17.  50  :  32  and  the  triplicate  of  4  :  5. 

18.  The  duplicate  of  x^ :  y"^  and  the  triplicate  of  y  :  x. 


X 

z 

y ' 

■  w 

a 

b 

b  ' 

'  c 

x' 

X' 

598  FIKST  COURSE  IN  ALGEBRA 

Which  is  the  greater  ratio  : 

19.    3  :  4  or  4  :  5  ?        20.    4  :  11  or  2  :  5  ?        21.    7  :  8  or  23  :  24 1 

Arrange  in  order  of  increasing  values  : 

22.    5  :  7,  6  :  8,  9  :  14,  and  27  :  28. 
Which  ratio  is  greater,  (1)  for  a;  positive,  (2)  for  x  negative: 
23.    2  :  3  or  (2  +  x)  :  (3  -\- x)%       24.    7  :  4  or  (7  -  a)  :  (4  -  cc)  ? 
Find  the  value  of  the  ratio  x\y\\i  each  of  the  following  : 
25.   ^l{^^-\-f)  =  V6{xy-f).     26.    10  (ar' +  /)  =  29  a^. 

II.  Propoetion 
16.   A  proportion  is  an  expressed  equality  of  two  equal  ratios. 

E.  g.  The  abstract  numbers  a,  6,  c,  and  d  are  said  to  be  in  proportion  in 
the  given  order  «,  6,  c,  </,  if  a  :  6  =  c  :  d  (read  "  a  is  to  6  as  c  is  to  d,  "). 

16.  The  numbers  a,  h,  c,  and  d  are  called  the  terms  of  the 
proportion  a  :  b  ^=  c  :  d. 

The  first  and  fourth  terms,  a  and  </,  are  called  the  extremes, 
and  the  second  and  third  terms,  b  and  c,  the  means,  of  the 
proportion. 

17.  In  a  proportion  a  \  b  =  c  :  d^  the  antecedents  a  and  c  of  the 
equal  ratios  a  :  b  and  c  :  d  are  called  the  antecedents  of  the 
proportion. 

18.  Similarly,  the  consequents  b  and  d  of  the  equal  ratios  are 
called  the  consequents  of  the  proportion. 

19.  In  any  proportion,  a  :  b  =  c  :  dj  the  value  of  either  of  the 
equal  ratios  is  called  the  common  ratio  of  the  proportion. 

20.  The  symbol  :  :  is  frequently  used  instead  of  the  equality  sign 
between  the  equal  ratios  of  a  proportion. 

E.  jT.    2  :  3  : :  4  :  6  instead  of  2  :  3  =  4  :  6. 

21.  The  numbers  a,  bj  c,d,  e,f, ,  are  said  to  be  in  contin- 
ued proportion  if  -  =  -  =  -  =  -  =  -  = ,  the   consequent 

0       c      a       e      J 

of  any  ratio  being  equal  to  the  antecedent  of  the  one  following. 

22.  In  a  continued  proportion,  a  :  b  =  b  :  c,  containing  three 
numbers,  a,  b^  and  c,  b  is  called  the  mean  proportional  between 
a  and  c,  and  c  is  called  the  third  proportional  to  a  and  b. 


PROPORTION  599 

23.  In  a  proportion,  a  .b  =  c  .  d,  containing  four  numbers,  a,  h, 
c,  and  (/,  the  fourth  number  d  is  called  the  fourth  proportional 
to  the  three  numbers  a,  b,  c,  in  the  given  order,  a,  b,  c. 

24.  Proportions  may  be  transformed  according  to  the  following 

General  Principles 

(i.)  In  any  numerical  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

That  is,  a  aih  =  cid,     then  ad  =  he. 

For,  multiplying  both  ratios  of  the  proportion  ajb  =  c/d  by  bd  we 
obtain  ad  =  be. 

This  principle  may  be  used  to  determine  whether  or  not  four 
numbers  are  proportional  in  some  specified  order. 

(ii.)  The  mean  projxyi'tional  between  any  two  numbers  is  equal  to 
the  square  root  of  their  product. 

That  is,  if  a  :  6  =  6  :  c,  (1),  then  h  —  ^sfac. 

For,  from  (1)  we  have  ^^  =  ac.     Hence  b  =  ^/ac. 

It  should  be  observed  that,  since  the  two  numbers  which  form 
either  of  the  ratios  of  a  given  proportion  must  be  numbers  of  the 
same  kind,  it  is  impossible  that  one  should  be  positive  and  the 
other  negative. 

Accordingly,  when  a  and  c  are  both  positive  numbers,  the  positive 
sign  only  is  taken  for  ^fac  in  the  expression  for  the  mean  propor- 
tional, b  =  \/«c.     (Compare  with  Chap.  XXVII.  §  39.) 

If,  however,  the   numbers   of  which  the  mean   proportional   is 
to  be  found  are   both   negative,  for  example  —  4  and  —  9,  we 

—  4:        m 
have,  representing  the  mean  proportional  by  m^ =  -— -  • 

Hence,  m  —  ^(  —  4)(  —  9)  =  —  6,  which  is  a  negative  number, 
(iii.)  If  the  product  of  two  numbers  is  equal  to  the  product  of  two 
others,  the  numbers  of  either  set  may  be  made  the  extremes,  and  those 
of  the  other  set  may  be  made  the  means  of  a  proportion. 

That  is,  if  ad  =  6c, 
we  obtain,  dividing  both  members  by  66?,   a  lb  =  cl  d. 

Therefore  either       €t\h  =  cid     (1)     or    cd  =  a:b.         (2) 


600  FIRST  COURSE   IN   ALGEBRA 


Similarly, 

Dividing  by  ac, 

d  :  c  =b  :  a     (3)     or     h  :  a  =  d\  c. 

(4) 

Dividing  by  ab, 

d  -.1  =  c  :  a     (5)     or     c  :  a  =  d  :  b. 

(6) 

Dividing  by  cd^ 

a  :  c  =  b  :  d     (7)     or     b  :  d  =  a  :  c. 

(8) 

It  appears  that  if  a,  b,  c,  and  d  form  a  proportion  in  any  one  of 
the  eight  orders  given  above,  they  also  form  a  proportion  when 
\vritten  in  any  one  of  the  remaining  orders. 

From  the  results  obtained  above,  it  may  be  seen  that  the  means 
of  a  proportion  may  be  interchanged.     Hence, 

(iv.)  if /our  numbers  or  quantities  of  the  same  kind  are  in  pro- 
portion^ the  terms  may  be  rearranged  by  alternation  ;  that  isy  tha 
first  term  is  to  the  third  term  as  the  second  term  is  to  the  fourth  term.. 

That  is,  if  a  :  6  =  c  :  <f ,  then  a  :  c  =  b  :  d. 

It  should  be  observed  that  the  terms  of  a  concrete  proportion  can  be 
transformed  by  alternation  only  when  the  terms  of  both  ratios  are  quantities 
of  the  same  kind.  For,  if  the  quantities  appearing  in  the  terms  of  the  first 
ratio  are  of  a  different  kind  from  those  appearing  in  the  terms  of  the  second 
ratio,  the  transformation  by  alternation  will  result  in  a  proportion  in 
which  the  terms  of  the  first  ratio  are  quantities  of  different  kinds  and  the 
terms  of  the  second  ratio  are  also  quantities  of  different  kinds. 

(v.)  In  any  proportion  the  terms  may  be  rearranged  by  inversion, 
that  isj  the  second  term  is  to  the  first  term  as  the  fourth  term  is  to  the 
third  term. 

That  is,  if  a  :  6  =  c  :  dy  then  hia  =  dic. 

For,  expressing  the  proportion  a  -.b  =  c  :  dm  the  fractional  no- 
tation, dividing  unity  by  each  member  and  simplifying,  we  obtain 
b/a  =  d/c. 

(vi.)  In  any  proportion^  the  terms  may  be  combined  by  addition; 
that  is,  the  sum  of  the  first  and  second  terms  is  to  either  the  first  term 
or  the  second  term  as  the  sum  of  the  third  and  fourth  terms  is  to 
either  the  third  term  or  the  fourth  term. 

That  is,  if  a  :  6  =  c  :  c«,  (1) 

then  {a  +  ft):  a  =  (c  +  d)i  c,  (2),  and  also  (a  +  6):  6  -  (c  +  d)i  d,  (8) 


PROPORTION 


601 


(3)  may  be  derived  as  follows  : 

Employing  the  fractional  nota- 
tion for  the  ratios,  we  may  express 
(1)  as  the  equation 
a  _  c 
h~d' 
Adding  unity  to  each  ratio, 


h^=2^- 


Hence, 


a-\-b 


c  +  d 
~d~' 


(2)  may  be  derived  as  follows  : 

Writing  (1)  by  inversion,  and 
expressing    the    result   in   frac- 
tional form,  we  obtain 
h_d, 
a       c 
Adding  unity  to  each  ratio, 


Hence, 


a 

a-\-b 


c 
c  +  d 


(vii.)  In  any  proportion  the  terms  may  be  combined  by  suhtrac- 
Hon ;  that  is^  the  difference  between  the  first  term  and  the  second 
term  is  to  either  the  first  term  or  the  second  term  as  the  difference 
between  the  third  term  and  fourth  term  is  to  either  the  third  term 
or  the  fourth  term. 

That  is,  if  a:b  =  c:  d,  (1) 

then  {a-b)ia={c-tl)i  c,  (2)  and  also  {a  -  h)  ih  =  {c  -  d) :  d,  (3) 

(3)  may  be  obtained  as  follows : 

Expressing  the  ratios   of  the 
given  proportion  by  the  fractional 
notation,  we  have 
a  _c 
b~d' 
Subtracting   unity  from    each 
ratio,  we  have 


a          _  c 
b~     ~d 


Hence, 


—  1. 

c  —  d 


(2)  may  be  obtained  as  follows: 

Writing  (1)  by  inversion  and 
expressing  the  result  in  fractional 
form,  we  have 

b  _d 
a      c 

Subtracting  both 
unity,  we  obtain 

a 

ratios  from 

d 

c 

TTpnr»A      =  

-d 

(viii.)  In  any  proportion  the  terms  may  be  combined  by  addition 
and  subtraction  ;  tJmt  is,  the  sum  of  the  first  and  second  terms  is 
to  their  difference  as  the  sum  of  the  third  and  fourth  terms  is  to 
their  difference. 

Thatis,  ifa:&=c  :  e«,  (1),  then  a  +  &  .  a-h^c  + d  :  c-d.   (2) 


602  FIRST  COURSE   IN  ALGEBRA 

The  proportion  (2)  may  be  obtained  by  dividing  the  ratios  ob- 
tained by  applying  (vi.)  to  (1)  by  the  corresponding  ratios  obtained 
by  applying  (vii.)  to  (1). 

(ix.)  In  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to 
the  sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 

That  is,  if  «  :  Z>  =  c  :  €?  =  e  :  /  =1 =  in  :  n, 

then  (a  +  c  +  €  + -\-  tn)  :  (h  +  d -\- f -\- +  n)  = 

a  :  b  =  c  :  d  =  e  :  f= =  ni  :  n^ 

provided  that  (b  +  d  -{■f+ -f  ??)  ^t  0. 

Let  r  denote  the  value  of  each  of  the  e(]ual  ratios. 

1  hen  rrom  7  =  r,      j  =  r,     >=  ^,     ,   —  =  r, 

0  a  J  n 

we  have  a  =  hr^   c  ^  dr,   e  ^=fr^ ,   m  •=^nr. 

By  addition,  (a  +  c  +  e  + +  m)  =  (ft  +  cZ  +/+ -{-n)r. 

Then  on  condition  that  (/>  +  ^  +  /  + +  «)  ^  0, 

,  a  +  f'  +  ^+ +  w  a      c  m 

we  have     .    ,    ,  .   ^  . ; —  =  ^  =  t=j= =— • 

(x.)  The  products  or  the  quotients  of  the  corresponding  terms  of 
two  projxyrtions  form  a  proportion. 

That  is,  if  a  :  6  =  c  :  r/,      (1),  and     oc  :  y  =  z  i  w,  (2), 

then  aoc  :  &//  =  cz  :  dw,  (8),  and  also  a/icih  I  y  =  c/zi  d/  iv»     (4) 

Writing  (1)  and  (2)  in  fractional  form,  and  multiplying  corre- 
sponding members  of  the  equations,  we  obtain  (3). 

Similarly,  (4)  is  obtained  by  division. 

(xi.)  Like  powers  o?-  like  principal  roots  of  the  terms  of  a  propor- 
tion are  in  proportion. 

That  is,  if  a:h  =  c:d,     (1) 

then     a** :  b''  =  c"  :  <^^    (2),  and  also  ^a:^b  =  ^c:  ^d.     (3) 

Let  a  :  b  =  r  and     c  :  d  =  r. 

Then  from  a  =  br^   (4)    and  c  =  dr,    (5) 

we  have  a"  =  ^"^•",         and         c"  =  6/"r". 

Hence  ^  ^*^'  ^°^        ^  ^  ^* 

Therefore  ^~^/   (^) 


PROPORTION  603 

Similarly,  from  (4)     and     (5), 

111  111 

al'  =  b'^r^y         and  c*  =  c?  V, 
1        1 

and  finally,  —  =  —  •  (7) 

l,k  flic 

Ex.  1.    Find  the  mean  proportional,  x,  between  4  and  25. 
Let  4  :  a;  =  a; :  25.     Then  x  =  ^^4  •  25  =  10. 

Ex.  2.    Show  that  if  a  :  6  =  c  :  c?  it  follows  that 

(a2  +  c2)  :  {ah  +  erf)  =  (tt&  +  cd)  :  (62  +  d^).  (1) 

Let  a  :  6  =  r  and  c  :  d  =  r. 

Then  a  =  &r     (2)     and  c  =  dr.     (3) 

Rence,  a^  +  c^  =  bh^  +  rfV.     (4) 
Multiplying  the  members  of  (2)  and  (3)  by  b  and  d  respectively  and 
adding  the  corresponding  members  of  the  resulting  equations,  we  have 

ab  +  cd  =  bh  +  dh.     (5) 
Fron.(4)a„d(5).  '^^  =  ^5^^^r.  (6) 

Similarly,  ____^  = -___^-  =  n  (7) 

Therefore,  (1)  follows  from  (6)  and  (7). 


Ex.3.   Solve  V£+lziV^  =  3  (1) 

/y/a:  +  6  +  ya;  -  15       ' 

Applying  0-iu.)  to  (1),  '^^^  =  -^  (2) 

—  2y  a:  —  15      ~  * 

a: +  6    _25 
ar-15~  4 
4  a:  +  24  =  25  a:  -  375 
a;  =19. 
Verifying  by  substituting  19  for  x  in  (1),  f  =  f  • 

Exercise  XXVI.    2 
Find  the  fourth  proportional  to 
1.  2,  3,  and  4.  3.  15,  16,  and  14. 

2-  14,  15,  and  16.  4.  16,  15,  and  14, 


604  FIRST  COURSE  IN  ALGEBRA 

Find  the  mean  proportional  between 

5.  2  and  8.  6.  1  and  4.  7.  12  and  6.  8.  aand-- 

a 
Find  the  third  proportional  to 

9.  2  and  8.  10.  5  and  10.  11.  ^  +  -  and  y  • 

b       a         0 

Construct  proportions   from   the  following  products  : 

12.  a'  =  be.  13.  x'  =  4-40.  14.  (a  +  b)(a  —  b)  =  c\ 

If  a:b  =  c  :d  obtain  each  of  the  following  proportions  : 

15.  a:b  =  ^:--      IQ.  ac  :bd  =  (^  :d'.    11.  a-{-c  :b+d=a'd -.bh. 
a    c 


18. 
19. 

(fl 
(a 

^by 
^by 

a' 
4  a' 

-5b' 

21. 
22. 

a  :b  = 
a'-b' 

'a  +  b' 

Vb''- 

-d":- 
c 

-d\ 

+  d' 

{c 

+  d)'- 

~  4  c' 

-5d' 

23. 

Ua:b 

=  b  :  c  show  that 

20. 

a\ 

a  +  c= 

a-\-b.a  +  b-[-c 

'.  +  d. 

a  +  b 

b-c 

.^  + 

c   a- 

-b 

a     '     b  b     '     a 

24.  What  quantity  must  be  added  to  the  terms  of  a^  -^  c^  to  make  it 
equal  to  a  :  c  / 

25.  What  expression  must  be  subtracted   from  each  of  the   following 
expressions  in  order  that  tlie  remainders  shall  form  a  proporti(m  ? 

4a4-6  +  f,     5a  +  6  +  c,     rt+13  6  +  c,     and     a+176  +  c. 

Solve  for  x  in  each  of  the  following  proportions  : 

26.  2:21  =  3:  x.  28.   (m'  -  n^)  :  {m  —  n)  =  x  :  1. 

27.  70  :  a.-  =  14  :  2.  29.  (a;  -  5)  :  3  =  5  :  12. 

Simplify  the   following  equations    by    applying   the   Principles 
Governing  Proportions,  and  then  solve  for  x  : 


30    "^"^  "^^  —  ^.  32    'V^  +  ^  +  vV 


V^  —  a/^      ^  a/^j  +  a  —  Va;  — 


a 


x+  ^/x—  \  _2l  ^/a-k-^a-\-x  _  Vc+Va;— c 

X  —  "s/x  —  I       1^  ya — \/a-\-x       Vc — 'Vxr—c 

34,  Find  three  numbers  in  continued  proportion  whose  sum  is  14  and 
whose  product  is  64. 

35.  Divide  $42.00  between  two  men  so  that  their  shares  shall  be  in  the 
ratio  of  3  :  4. 


PROBLEMS  IN  PHYSICS  605 

36.  Divide  44  into  two  parts  such  that  the  less,  increased  by  one,  shall 
be  to  the  greater,  decreased  by  one,  as  5  :  6. 

37.  Two  numbers  are  in  the  ratio  of  4  :  5.  If  each  is  increased  by  5  the 
sums  will  be  in  the  ratio  of  5  :  6.     What  are  the  numbers  ? 

38.  What  number  must  be  added  to  the  numbers  3,  4,  7,  and  9  in  order 
that  their  sums  shall  form  a  proportion  ? 

39.  What  number  must  be  subtracted  from  24,  27,  40,  and  55,  in  order 
that  tlie  remainders  shall  form  a  proportion  I 

40.  Find  the  ratio  of  the  numerator  of  a  fraction  to  its  denominator  if 
the  value  of  the  fraction  remains  unchanged  when  the  numerator  is  in- 
creased by  a  and  the  denominator  is  increased  by  b  ? 

The  areas  of  two  similar  plane  figures  have  the  same  ratio  as  the  squares 
of  any  two  corresponding  dimensions. 

41.  The  area  of  a  triangle  is  90  square  inches  and  tlie  base  is  12  inches. 
What  is  the  area  of  a  similar  triangle,  provided  that  the  base  is  16  inches? 

42.  The  area  of  the  first  of  two  simihir  polygons  is  128  square  inches,  and 
the  area  of  the  second  is  200  square  inches.  If  one  side  of  the  first  polygon 
is  8  inches,  find  the  corresponding  side  of  the  second  polygon. 

The  volumes  of  two  similar  solids  have  the  same  ratio  as  the  cubes  of 
any  two  corresponding  dimensions. 

43.  The  diameter  of  the  first  of  two  bottles  which  are  of  similar  shape  is 
three  times  that  of  tlie  second.  If  the  first  holds  2  ounces,  how  much  does 
the  second  hold  ? 

44.  If  a  sphere  which  is  2  inches  in  diameter  weighs  5  lbs.,  what  is  the 
weight  of  a  sphere  of  the  same  substance  which  is  3  inches  in  diameter  ? 

Problems  in  Physics 

25.  The  Inclined  Plane.  If  a  bodt/  rests  on  a  smooth  in- 
clined plane,  the  force  {disregarding  friction)  which  must  be  applied 
along  tJie  plarie  to  hold  the  body  in  place  against  the  action  of  the 
force  of  gravity  has  the  same  ratio  to  the  weight  of  the  body  that  the 
height  of  the  plane  has  to  the  length  of  the  plane. 

26.  If  F  represents  the  force  applied  along  an  inclined  plane, 
W  the  weight  of  the  body,  h  the  height  of  the  plane,  and  /  the 
length  of  the  plane,  we  have 

W      l' 

The  force  represented  by  F  which  is  applied  along  the  plane  is 
called  the  component  of  the  weight  W  which  is  parallel  to  the  plane. 


606  FIRST  COURSE  IN  ALGEBRA 

Exercise  XXVL    3 
Solve  the  following  problems : 

1.  Find  the  force  which  must  be  exerted  to  draw  a  sled  weisrhinsr 
240  lbs.  up  a  hill  which  is  300  feet  long  and  50  feet  high. 

2.  What  is  the  weight  of  a  body  if  a  force  of  125  lbs.,  exerted  along  a 
smooth  inclined  plane  which  is  80  feet  in  length  and  20  feet  in  height, 
prevents  the  body  from  sliding  down  the  plane? 

3.  A  boy  who  is  able  to  exert  a  maximum  force  of  80  lbs.  is  able  to 
keep  a  barrel  from  rolling  down  a  plank  which  is  12  feet  in  length  and 
the  upper  end  of  which  is  3  feet  from  the  ground.  Find  the  weight  of 
the  barrel. 

4.  A  porter  who  can  exert  a  maximum  force  of  200  lbs.  undertakes  to 
roll  a  cask  weighing  500  lbs.  up  a  board  wliich  is  10  feet  long.  How  high 
can  the  u])per  end  of  the  board  be  placed  without  compelling  the  porter  to 
allow  the  cask  to  roll  down  the  board  ? 

5.  A  car  weighing  1200  lbs.  is  held  at  rest  on  a  smooth  inclined  plane 
by  a  force  of  30  lbs.  applied  parallel  to  the  plane.  If  the  length  of  the 
plane  is  800  feet,  find  the  height  of  the  plane. 

6.  A  boy  is  able  to  exert  a  maximum  force  of  80  lbs.  How  long  an 
inclined  plane  must  he  use  to  push  a  truck  weighing  320  lbs.  up  to  a 
doorway  which  is  3^  feet  above  the  ground  1 

Boyle's  Law.  Tlie  volume  of  a  gas  is  {approximately)  inversely 
proportional  to  thk  pressure^  providsd^  that  the  temperature  remains 
constant. 

That  is,  representing  the  pressure  by  Pi  when  the  volume  is  Fi, 
and  the  pressure  by  Pi  when  the  volume  is  V^^  we  have 

V,      Pi 

7.  If,  when  confined  with  a  pressure  of  20  lbs.  per  square  inch,  a  mass 
of  gas  occupies  a  volume  of  one  cubic  foot,  find  the  volume  of  the  gas 
when  the  pressure  becomes  40  lbs.  per  square  inch. 

8.  A  gas  bag  containing  3  cubic  feet  of  gas  under  a  pressure  of  18  lbs. 
per  square  inch  must  be  subjected  to  what  pressure  to  reduce  the  volume  to 
half  a  cubic  foot  1 

9.  Six  cubic  feet  of  gas  under  a  pressure  of  45  lbs.  per  square  inch  will 
have  what  volume  if  the  pressure  is  reduced  to  15  lbs.  per  square  inch  ? 

10.   A  bladder  holds  40  cubic  inches  of  air  under  a  pressure  of  15  lbs. 


^PROBLEMS   IN   PHYSICS  607 

per  square  inch.     What  is  the  size  of  the  bladder  when  the  pressure  is 
reduced  to  12  lbs.  per  square  inch  ? 

11.  When  under  a  pressure  of  75  lbs.  per  square  inch,  the  volume  of  a 
a  mass  of  gas  is  128  cubic  inches.  What  is  the  pressure  when  the  volume 
becomes  240  cubic  inches  ? 

12.  If  100  cubic  inches  of  air,  at  a  pressure  of  27  lbs.  per  square  inch, 
be  admitted  to  a  vessel  the  volume  of  which  is  450  cubic  inches,  what  will 
be  the  pressure  i 

By  absolute  temperature  expressed  in  degrees  centigrade  is 
meant  the  number  of  degrees  above  0°  C,  plus  273°  C 

Representing  the  absolute  temperature  by  T^  and  the  number 
of  degrees  above  0°  C.  by  tj  we  have 

T  =t  +  273. 

Charles's  Law.  The  volume  of  a  gas  is  directly  proportional  to 
the  absolute  temperature^  provided  that  the  pressure  remains  constant. 

Assuming  that  the  pressure  remains  constant,  we  will  repre- 
sent by  Vi  the  volume  of  a  mass  of  gas  when  the  absolute  tem- 
perature is  7\°  C,  and  by  Fg  the  volume  when  the  tempera- 
ture is  2^2°  C.     Then  we  have  (approximately) 

13.  The  volume  of  a  certain  quantity  of  gas  is  100  cubic  centimeters  at 
0°  C.  At  what  temperature  will  the  volume  become  200  cubic  centimeters, 
assuming  that  the  pressure  remains  constant  ? 

14.  If  the  volume  of  a  certain  mass  of  gas  is  500  cubic  centimeters  at 
20°  C,  find  the  volume  of  the  gas  at  87°  C,  assuming  that  the  pressure 
remains  constant. 

15.  A  mass  of  gas  occupying  a  volume  of  160  cubic  centimeters  at  a 
temperature  of  47°  C.  is  cooled  to  a  temperature  of  17°  C.  Find  the  volume 
at  the  lower  temperature. 

16.  A  certain  mass  of  gas  occupying  a  volume  of  90  cubic  centimeters  at 
12°  C.  is  raised  in  temperature  to  50°  C.  Find  the  volume  at  the  higher 
temperature. 

Representing  by  Fi  the  volume  of  a  gas  when  the  pressure  is  P\ 
and  the  absolute  temperature  is  Ji°  C,  and  by  V^  the  volume  of 
the  gas  when  the  pressure  is  P^  and  the  absolute  temperature  of 
7^2°  C,  it  may  be  shown  that  the  following  relation  is  true: 

VxP.  _  V2P2 


608  FIRST  COURSE  IN  ALGEBRA 

Since  the  barometer  is  commonly  used  to  measure  the  pressure 
of  a  gas,  it  will  be  convenient  to  give  the  pressures  of  the  gases 
in  the  following  examples  in  terms  of  the  height  of  a  column  of 
mercury. 

17.  Five  hundred  cubic  centimeters  of  a  gas  at  a  temperaturo  of  27°  C 
are  cooled  to  2°  C,  and  at  the  same  time  tlie  external  pressure  upon  the 
gas  is  changed  from  74  centimeters  of  mercury  to  76  centimeters  of  mercury. 
What  does  the  volume  of  the  gas  become  ? 

18.  Fifty  liters  of  gas  are  generated  at  a  temperature  of  12°  C-.  and  a 
pressure  of  68  centimeters  of  mercury.  Find  the  volume  of  the  gas  at  ()°  0. 
when  the  pressure  is  76  centimeters  of  mercury. 

19.  The  volume  of  a  certain  quantity  of  <^'as  is  fouu<l  to  be  300  cubic 
centimeters  at  a  temperature  of  0°  C.  and  pressure  of  75  centimeters  of 
mercury.  What  must  be  the  temperature  of  the  <j;as  in  order  tliat  the 
volume  may  be  350  cubic  centimeters  when  the  pressure  is  76  centimeters 
of  mercury  ? 

20.  The  volume  of  a  certain  mass  of  gas  is  found  to  be  784  cubic  centi- 
meters at  a  pressure  of  75  centimetei*s  of  mercury  and  a  temperature  of  7°  C. 
What  nmst  be  the  pressure  in  centimeters  of  mercury  if  the  volume  of  the 
gas  becomes  900  cubic  centimeters  when,  the  temperature  is  27°  C,  ? 

Linear  Expansion  of  Solids.  A  solid  expands  when  heated, 
and  the  increase  in  length  of  a  substance  in  the  form  of  a  bar  is 
approximately  proportional  to  the  original  length  of  the  bar  and  to 
the  change  in  temperature,  provided  that  the  change  in  temperature 
is  small. 

The  coefficient  of  lineai'  expansion  of  a  body  is  the  ratio  of  the 
increase  in  length  per  degree  rise  in  temperature  to  the  length  of 
the  body  at  0°  C. 

If  k  represents  the  coefficient  of  linear  expansion  of  a  bar  the 
length  of  which  at  0°  C.  is  represented  by  4  and  the  length  of 
which  at  tx°  C.  is  represented  by  /i,  we  have 

/i  =  /o  +  hkh  =  4  (1  +  ^^i)- 
If  the  length  of  the  bar  at  #2°  C.  is  represented  by  4,  we  have 

Hence,  from  the  relations  above,  we  have  the  following  proportion : 
/,  _  1  +  kh 
Ur  1  -Vkt^' 


VARIATION  609 

21.  An  iron  steam  pipe  is  75  feet  in  length  at  0°  C.  What  does  its 
length  become  when  steam  at  a  temperature  of  112°  C.  is  passed  through  it, 
provided  that  the  coefficient  of  linear  expansion  of  iron  is  .000012  ? 

22.  The  distance  between  two  graduations  on  a  brass  bar  is  exactly  one 
meter  at  25°  C.  What  is  the  distance  between  the  graduations  at  60°  C, 
provided  that  the  coefficient  of  linear  expansion  of  brass  is  .000018  ? 

23.  A  lightning  rod  which  is  made  of  copper  is  40  feet  in  length  when 
at  a  temperature  of  0°  C.  Find  the  length  of  the  rod  when  the  tempera- 
ture is  30°  C,  provided  that  the  coefficient  of  linear  expansion  of  copper 
is  .000017. 

24.  What  allowance  should  be  made  for  expansion  in  a  1700-foot  span  of 
a  steel  bridge,  assuming  that  the  highest  summer  temperature  is  40°  C.  and 
the  lowest  winter  temperature  is  20°  C,  provided  that  the  coefficient  of 
linear  expansion  of  steel  is  .00001 1  ? 

25.  If  a  steel  rule  is  exactly  one  foot  in  length  at  a  temperature  of  0°  C, 
find  the  error  in  the  rule,  expressed  as  a  fraction  of*  an  inch,  at  a  tem- 
perature at  20°  C,  provided  that  the  coefficient  of  linear  expansion  of  steel 
is  .000011. 

III.  Variation 

27.  One  variable  number  or  quantity  is  said  to  be  a  function 
of  a  second  if  a  change  in  the  value  of  the  second  produces  in  general 
a  change  in  the  value  of  the  first.     (See  also  Chap.  IX.    §  1.) 

If  the  values  of  two  variables  are  so  related  that  the  value  of  the 
first  variable  is  regarded  as  depending  upon  the  value  which  may 
be  assigned  to  the  second  variable,  then  the  first  is  called  the 
dependent  variable  and  the  second  the  independent  variable. 

E.  g.  If  the  values  of  x  and  y  be  restricted  to  satisfy  the  conditional 
equation,  x  =  y  -{•  7,  the  value  of  either  variable  depends  upon  the  value 
which  may  be  assigned  to  the  other. 

If  y  be  selected  as  the  independent  variable,  then  by  assigning  succes- 
sively the  values  0,  1,  2,  3,  4,  etc.,  to  y,  the  dependent  variable  x  will  assume 
successively  the  values  7,  8,  9,  10,  11,  etc. 

One  Independent  Variable 

28.  Two  variable  numbers  or  quantities  are  said  to  vary  directly 

one  as  the  other,  if  when  the  value  of  one  is  changed  the  value  of 
the  other  is  changed  also,  and  in  the  same  ratio. 

E.  g.  The  distance  travelled  in  an  hour  by  a  person  walking  uniformly 
varies  directly  as  the  rate. 

39 


610  FIRST  COURSE   IN  ALGEBRA 

29.  The  symbol  of  direct  variation,  qc,  (which  is  read  "varies 
directly  as,"  or  simply  "varies  as"),  when  placed  between  two 
variables  denotes  that  their  ratio  is  a  constant  number. 

E.g.   Representing  some  constant  number  by  c,  we  may  from  a;  oc  y, 

obtain  -  =  c. 

y 

30.  As  a  result  of  the  definition  above,  we  have  the  following 
Fuudaniental  Principle :  1/  the  value  of  x  depends  on  the  value 
of  y  and  y  alone  in  such  a  way  that  if  (ic,  y)  and  (a;i,  y^  repi^e- 
sent  two  pairs  of  corresponding  values^  and  if  for  every  two  stick 
pairs  we  have  x  :  Xi  =  y  :  yi,  then  it  follows  that  x  is  a  constant 
multiple  of  y. 

For,  if  -  =  ^,  (1) 

xi      yx  \^ 

it  follows  that  ?=^.  (2) 

y      yi 

If  we  let  Xj  an<l  y^  denote  any  particular  pair  of  corresponding  values  of 
X  and  y,  while  x  and  y  denote  any  other  pair  of  corresponding  values,  it 
appears  that,  if  the  ratio  of  all  pairs  of  values  are  equal  to  the  ratio  of  the  • 
same  pair,  x^  and  j/i,  as  in  (2),  we  may  denote  the  value  of  the  ratio  of  this 
particular  pair  chosen  for  reference  by  some  constant,  c. 

Hence  (2)  becomes  xjy  =  c  (3),  ot  x  =  cy  (4),  in  which  c  denotes  some 
constant. 

31.  Since  the  ratio  of  any  pair  of  corresponding  values,  x  and  y, 
is  equal  to  the  ratio  of  any  other  pair  of  corresponding  values,  it 
follows  that  any  two  pairs  of  corresponding  values  may  be  used  to  , 
form  a  proportion. 

Accordingly  x  ^  y  \&  often  read,  "a;  is  proportional  to  y^ 

32.  From  x  =  cy,  it  follows  that  the  value  of  the  constant  c  may 
be  obtained,  provided  that  the  value  of  x  corresponding  to  any 
specified  value  of  y  is  known. 

Ex.  1.    \i  X  (X.  y  and  a;  =  6  when  y  =  3,  find  x  when  y  =  13. 
From  X  ac  y  we  obtain  the  conditional  equation  x  =  cy,  in  which  c  de- 
notes some  constant. 

Substituting  6  and  3  for  x  and  y  respectively,  we  obtain 
6  =  c-3     or     c  =  2.         Hence     x  =  2y. 
Accordingly,  when  y  =  13,  we  have  x  =  2  •  13  or  26. 


VARIATION  611 

33.  One  variable  number  or  quaatity  is  said  to  vary  inversely 
as  a  second,  or  to  be  inversely  'proportional  to  a  second^  when  the 
first  number  varies  as  the  reciprocal  of  the  second. 

We  may  indicate  that  x  varies  inversely  as  3/,  or  is  inversely  pro- 
portional to  ?/,  by  writing  a;  «  -  • 

1  X  .  . 

From  the  notation  aj  a  -  we  have  -r-  =  c,  in  which  c  denotes 

y  1 

y 

some  constant.     Hence  xy=-c. 

E.  g.  If  eight  men  can  do  a  given  amount  of  work  in  24  hours,  hy  reduc- 
ing the  number  of  men  to  four,  the  time  required  to  do  the  same  work 
would  be  increased  to  48  houra  ;  two  men  would  require  96  hours,  and  one 
man  alone  192  hours. 

It  may  be  seen  that  the  products  8  x  24,  4  x  48,  2  x  96,  and  1  x  192, 
are  all  equal. 

Two  or  More  Independent  Variables 

34.  One  variable  number  or  quantity  is  said  to  vary  as  two 
others  jointly,  or  to  be  proportional  to  two  others  jointly ^  if  it 
varies  as  the  product  of  the  other  two. 

We   may  indicate  that  x  varies  as  y  and  z  jointly  by  writing 

X  oc  yz,  from  which  it  follows  that  —  =  c,  in  which  c  denotes  a 

yz 

constant. 

E.  g.  The  distance  passed  over  by  a  body  moving  uniformly  is  propor- 
tional both  to  the  rate  and  the  time. 

35.  One  variable  number  or  quantity  is  said  to  vary  directly 
as  a  second  and  inversely  as  a  third  if  it  varies  directly  as  the 
second  and  inversely  as  the  third  jointly. 

We  may  indicate  that  x  varies  directly  as  y  and  inversely  as  z  by 

writing  x  ^  -t  from  which  we  obtain  —  =  c,  in  which  c  denotes 

^       z  y 

z 

x^ 
some  constant.     Accordingly  ~  =  c. 

E.  g.  The  time  required  to  complete  a  journey  varies  directly  as  the 
distance  and  inversely  as  the  rate. 


61^  FIRST  COURSE  IN  ALGEBRA 

36.    From  the  definitions  above  we  obtain  directly  the  following 
General  Principles 

In  the  proofs  of  the  following  principles,  a;,  y^  z^  etc.,  represent 
variables,  and  c,  Ci,  etc.,  represent  constants. 

(i.)  Ifx<x.  y  and  y  ^  Zj  then  ic  ^  z. 

For,  i{x<x  y  then  -  =  c,  and  if  y  cc  z,  then  -  =  Ci. 

y  z 

Therefore    -  x  -  =  cci,     or    -  =  cci.        Hence,     x  ^  z. 
y      z  '  z 

(ii.)    If  ac  c^  z  and  y  oc  tf ,  then  xy  a  zw. 

For,  \ix  ^  Zy  then  7  =  f ,  and  if  ?/  «  ?^;,  then  —  =  ci. 

Therefore,     -  X  —  =  Wi,     or    -^  =  cci :  that  is,  xy  oc  ;2;2^. 

;3        W  ZW  ^ 

(iii.)   If  oc  cc  yz,  then  x/y  oc  s,  anrf  x/z  oc  //. 

For,  if  a;  oc  yz^  then  —  =  c. 

a? 

X  ~~  11  11 

Hence     '-^  =  c,     or    —  =  c :  that  is,     xl  y  oc  z. 

Similarly    xl  z  ^  y. 

(iv.)    If  X  ^  z  and  y  oc  «,  «/*c»  x  -{-  y  ct:  z  and x-y  ^  z. 

X  Ii 

For,  if  a;  oc  ;:;,  then  -  =  c,  and  if  ;y  oc  5,  then  ~  =  c. 


Therefore       -  +  ^  =  c  +  Cj. 


2; 

X 


Also, -^  =  c  —  Ci. 

;2        Z 

That  is, ^  =  c  —  Ci. 

Hence,  x  —  y  ^  z. 


That  is,  ^-±^  =  c  +  ci. 

z 

Hence,  x  +  y  ^  z. 

(v.)  iy  ^^e  'ya/?^^  o/a;  depends  upon  the  values  of  both  yand  z^  and 
on  these  alone,  and  ifx  oc  y  vjhen  z  is  constant  and  x  ^  z  when  y  is 
constant,  then  x  ^  yz  when  both  y  and  z  vary. 

To  establish  this  principle  we  will  suppose  that  (a^i,  yi,  Zi), 
(x,  yo,  Zi),  and  (xo,  yz,  -2)  represent  three  sets  of  corresponding 
values  of  the  variables  x,  y,  and  ;::.  These  values  are  such  that  in 
passing  from  the  first  set  to  the  second  set  the  value  of  z,  repre- 


VARIATION  613 

sen  ted  by  zi,  remains  constant,  and  in  passing  from  the  second  set 

to  the  third  set  the  value  of  ^,  represented  by  1/2,  remains  constant. 

Then  from  (xi,  1/1,  zi)  and  (x,  1/2,  zi),  since  Zi  remains  constant, 

we  have,  _i=r^.  (1) 

Also  from  (x,  1/2,  zi)  and  (o^,  1/2,  Z2),  since  2/2  remains  constant, 

we  have,  _  =  ^ .  (2) 

Xo         Z2  ^   ^ 

Accordingly,  multiplying  together  the  corresponding  members  of 
(1)  and  (2),  we  have 

ici  _  yxZx  ^ 

X2      yiZ2 

Hence,  = 

y\Z\      y2Z2 

Therefore  the  corresponding  values  of  {x^,  y^,  z^  and  (a^,  y^^  Z2) 
are  proportional ;  that  is,  x<xyz. 

The  principle  may  be  shown  to  apply  when  x  depends  for  its 
value  upon  the  values  of  three  or  more  variables,  y^  z,  w,  •  -  -  . 

Ex.  2.  If  X  QC  y  and  a;  =  20  when  y  =  4,  find  x  when  y  =  7. 

If  X  cc  y  we  may  assume  x  =  cy,  (I),  in  which  c  is  a  constant.  Since 
this  equation  is  satisfied  when  a;  =  20  and  y  =  4,  we  have  20  =  4  c.  There- 
fore c  =  5. 

To  find  X  when  y  =  7,  we  may  substitute  5  for  c  and  7  for  y  in  equation 
(1),  and  obtain  x  =  36. 

Ex.  3.  If  w  varies  as  x  and  y  jointly,  and  w  =  42  when  x  =  2  and  y  =  3, 
find  w  when  x  =  4  and  y  =  6. 

From  the  given  conditions  we  may  assume  that  iv  =  cxy,  (1)  in  which  c 
denotes  some  constant. 

Substituting  the  given  values  for  w,  x  and  y  in  (1),  we  find  that  c  =  7. 

We  may  substitute  7,  4,  and  5  for  c,  x,  and  y,  respectively,  and  obtain 
w  =  140. 

Exercise  XXVL    4 

1.  If  X  (X  y  and  when  y  =  8,  x  z=  56,  find  x  when  y  =  I. 

2.  If  a;  oc  l/y  and  a:  =  6  when  y  =  2,  find  x  when  ^  =  9. 

3.  If  a;  oc  l/y  and  a;  =  1  /2  when  y  =  16,  find  y  when  a:  =  2. 

4.  If  X  varies  jointly  as  y  and  z,  and  a:  =  24  when  y  =  4  and  s  =  2,  find 
x  when  ^  =  5  and  2!  =  3. 


614  FIRST  COURSE  IN  ALGEBRA 

5.  If  X  varies  directly  as  y  and  inversely  as  s,  and  a:  =  22  when  t/  =  1 1 
and  2  =  7,  find  x  when  y  =  36  and  2  =  9. 

6.  Find  x  when  y  =  3,  if  a:  oc  \/y^  and  a:  =  9  when  y  =  2. 

7.  Find  y  when  x  =  15  and  ty  =  3,  if  2/  varies  as  x  and  ly  jointly,  and 
y  =  1  when  x  =  12  and  «?  =  1 ,  8. 

8.  If  a;  oc  y,  show   that  ax  x  ay  when  a  is  either  a  constant  or  a 
variable. 

9.  Ifx  cc  l/y  and  y  cc  l/z  show  that  x  cc  z. 

The  area  of  a  circle  varies  as  the  square  of  its  diameter. 

10.  If  the  area  of  a  circle  the  radius  of  which  is  14  feet  is  616  square 
feet,  find  the  area  of  a  circle  the  radius  of  which  is  18  feet. 

11.  Show  that  the  area  of  a  circle,  the  diameter  of  which  is  10  inches, 
is  equal  to  the  sum  of  the  areas  of  two  circles,  the  diameters  of  which  are 
8  inches  and  6  inches  respectively. 

12.  The  volume  of  a  sphere  varies  as  the  cube  of  its  radius,  and  the 
volume  of  a  sphere  of  which  the  radius  is  3  inches  is  113y  cubic  inches. 
Find  the  volume  of  a  sphere  the  radius  of  which  is  5  inches. 

13.  Prove  that  the  sum  of  the  volumes  of  three  spheres,  the  radii  of 
which  are  3,  4,  and  5  inches  respectively,  is  equivalent  to  the  volume  of  a 
sphere  the  radius  of  which  is  6  inches. 

Problems  in  Physics 

It  has  been  found  by  experiment  that  the  distance  passed  over 
by  a  falling  body,  moving  freely  and  receiving  no  initial  impulse, 
varies  directly  as  the  square  of  the  time. 

14.  If  a  body  falls  16  feet  in  1  second,  how  far  will  it  fall  in  8  seconds  ? 

15.  A  stone  is  dropped  from  the  top  of  a  cliff  and  strikes  the  bottom  of 
the  cliff  in  3^  seconds,  nearly.  What  is  the  approximate  height  of  the 
cliff? 

16.  From  what  height  must  a  body  fall  from  a  state  of  rest  to  reach  the 
earth  after  10  seconds  ? 

It  has  been  found  by  experiment  that  the  velocity  acquired  by  a 
body  falling  freely  from  a  state  of  rest  varies  directly  as  the  time. 

17.  If  the  velocity  of  a  falling  body  is  180  feet  per  second  at  the  end  of 
5  seconds,  what  will  be  its  velocity  at  the  end  of  9  seconds  ? 

18.  If  the  velocity  of  a  falling  body  is  128  feet  per  second  at  the  end  of 
4  seconds,  what  will  be  its  velocity  at  the  end  of  7  seconds  ? 

The  intensity  of  illumination  from  a  source  of  light  varies  in- 
versely as  the  square  of  the  distance  from  the  source. 


PROBLEMS   IN   PHYSICS  615 

19.  A  candle  is  placed  at  a  distance  of  1  foot  from  a  cardboard  screen, 
and  a  second  candle  is  placed  at  a  distance  of  7  feet  from  the  screen  on  the 
other  side.  Compare  the  intensity  of  illumination  on  the  two  sides  of  the 
screen. 

20.  A  gas  jet  which  is  16  feet  from  a  photometer,  and  a  candle  which  is 
4  feet  from  the  photometer,  are  found  to  illuminate  it  equally.  Compare 
the  intensity  of  lij,dit  from  the  two  sources. 

21.  A  "standard"  16-caiidle-power  lamp,  when  placed  at  a  distance  of 
51  centimeters  from  a  screen,  is  found  to  illuminate  it  with  the  same  inten- 
sity as  an  incandescent  light  placed  at  a  distance  of  49  centimeters  from 
the  screen.     What  is  the  candle  power  of  the  incandescent  light  ? 

When  an  elastic  body  is  stretched,  it  is  found  that  within  the 
limits  of  perfect  elasticity  the  elongations  of  the  body  are  directly 
proportional  to  the  forces  producing  them. 

The  elongation  E  produced  by  a  stretching  force  F  upon  a 
substance  in  the  form  of  a  rod  of  diameter  D  and  length  X,  varies 
directly  as  the  force  F,  directly  as  the  length  Z,  and  inversely 
as  the  cross  section,  —  that  is,  inversely  as  the  square  of  the 
diameter  D. 

FL 
D' 

22.  If  a  certain  wire,  1/10  of  an  inch  in  diameter  and  36  inches  in 
length,  stretches  3  inches  under  a  force  of  18  lbs.,  how  much  will  it  stretch 
under  a  force  of  24  lbs  ? 

23.  A  certain  wire,  the  diameter  of  which  is  1/10  of  an  inch  and  the 
length  of  which  is  5  feet,  is  increased  in  length  4  inches  by  a  force  of  24  lbs. 
Find  the  length  of  a  second  wire  of  the  same  material  and  diameter  if  a 
force  of  40  lbs.  increases  it  in  length  by  7  inches. 

24.  If  a  wire  which  is  1/16  of  an  inch  in  diameter  and  25  feet  in  length 
stretches  3  inches  under  a  force  of  15  lbs.,  how  long  is  a  wire  the  diameter 
of  which  is  1  /20  of  an  inch,  if  a  force  of  40  lbs.  produces  an  increase  in 
length  of  2  inches? 


That  is,  ^«  n2 


Mental  Exercise  XXVI. 

5. 

Review 

Solve  each  of  the  following  equations ; 

1.  (x-Q>y=  16.                           4. 

2.  {x-^y  =  4.x',                          5. 

3.  a; +  1  =  9  +  ^-                            6. 

--Ty 

=  0. 
=  0. 

=  n. 

616 


FIRST  COURSE   IN  ALGEBRA 


7.  Show  that  a;'  —  1  =  0,  if  ic  —  1  =  0. 

8.  Show  that  a;'  —  4  =  0,  if  a;  —  2  =  0. 

9.  Show  that  a;-  —  9  =  0,  if  a  +  3  =  0. 

10.  Show  that  a^  —  ab  +  b""  =  0,  if  a  +  b  =  0. 

Simplify  each  of  the  following: 

Va-^  b  13.  (3  +  a/^)(3  -  ^/^^). 

14.    (2  +  V^){2  -V^. 


11 


12. 


V«  +  Vb 


15.  (2\/-3  -  3a/^2^ 

16.  (5V^^  -  2a/^^)^ 


Distinguish  between 

17.   —  2  ~  V^±_     and     —    4  ~  aA^- 
.18.   —5-7-  V—  25     and     —  25  -=-  V—  5. 

19.  Simplify       V250^';     \w^2>     V^^^^- 
Express  the  following  as  entire  surds: 

20.  2\^.  22.  3a;v^ 


23.  I  VSa 


a. 


25.  ^^8c. 
2c 


21.  a^^''. 

Simplify  and  express  with  positive  exponents : 

Solve  each  of  the  following  equations : 
31.  x^  =  9. 

30.  a;^=:4.  ^'>    -^ 


29    ^-^ 
^^-  3  ~  x' 


32.  a;^  =  1. 


33.  x^  =  -h 
34.^=1. 


35.   Rationalize  the  denominators  of    —iz and    — = 

V  a  +  1  V  a^  —  1 


From  each  of  the  following  conditional  equations  find  the  ratio 
ofxtoy: 


36.  2a;  =  3y. 


37.  5a;  =  2y. 


38.  7i/  =  4.x. 


REVIEW  617 

S9.Sx  =  2/.  40.  ic  =  |.  4:l.x  =  —  y. 

o  11 

Distinguish  between 

42.  I  +  -  and  a^  +  b^.  43.   (x  -  i/y  and  a;"^  -  2/-K 

Find  the  value  of  each  of  the  following  expressions  : 


1 


50.  3-2-22. 

51.  2-2. 


«.^.  «(!)-.  (I)--..-,.. 


53.  2-2  -  21 

54.  3-^-3. 


55.  (V-  2)«  +  (-V2)\ 
Show  that 

56.  12^  •  82  =  2^2  .  38.  57    g8  .  jg5  ^  28 .  3i». 

Simplify  each  of  the  following : 


V 


58.  a^-ira''.  65.  (« -^  a")". 

59.  x^  -^  ici  - .  7  a^» 

60.  a° 


f^    -I  5«*»  +  26<^ 


61.  ^.  67 

62.  4_. 


63.  -37=^-  69.  {Va  -  V^^Y 


64.  -jj:^'  70.  J- 

Find  the  mean  proportional  between 

71.  am^    and    am.  72.  6'c    and    bd^. 


618  FIRST  COURSE  IN   ALGEBRA 

Simplify  the  following  ratios  : 

73.  2/3  :  4/5.  77.  ajb  :  c/6. 

74.  3/7  :  3/8.  78.  xjy  :  z/x. 

75.  5/6  :  11/6.  79.  a/x  :  b/x. 

76.  3/4  :  4/3.  80.  m/n  :  n/q. 

Express  the  following  proportions  in  the  least  numbers  possible 
without  altering  the  terms  containing  x : 

^^•io-*  *'^- 10-75  '^^•eo-io 

15      x  36       24  00       o 

83.?2  =  !5.  87.1«  =  i?.  91.  g  =  ^. 

4        a;  15       jc  24       aj 

^^•42~6'  ^^•21~60  ^^•80"24 


THE  PROGRESSIONS  619 


CHAPTER  XXVII 

THE   PROGRESSIONS 

1.  A  SUCCESSION  of  numbers,  each  of  which  is  formed  according 
to  some  definite  law,  is  called  a  sequence. 

The  successive  numbers  are  called  the  terms  of  the  sequence. 

2.  It  should  be  understood  that  we  cannot  take  numbers  at 
random  to  form  a  sequence. 

There  must  be  some  definite  relation  between  the  numbers  chosen 
such  that  when  the  number  of  any  particular  term  of  the  sequence 
is  known,  its  value  can  be  computed. 

Hence  it  is  possible  to  determine  whether  or  not  any  specified 
number  occurs  in  a  given  seciuence. 

E.g.  1,2,3,4, ,  w, 

1,4,9,  16, ,n^, 

1/2,2/3,3/4, ,n/(n+l), 

3.  A  sequence  of  numbers,  ai,  a^,  aa,  a^y ,  n^„, ,  is 

said  to  be  given  or  known  if  the  value  of  any  specified  term  is 
known  or  can  be  found  when  the  position  of  the  term  in  the 
sequence  is  given. 

4.  The  law  governing  the  formation  of  the  successive  terms  may 
be  such  that  any  term  after  the  first  may  be  obtained  by  performing 
some  definite  operation  upon  the  term  which  immediately  precedes  it, 

E.  g.    In  the  sequence  5,  8,  11,  14, ,  each  term  is  obtained  by 

adding  3  to  the  next  preceding  term. 

In  the  sequence  2,  6,  18,  54,  162, ,  each  term  is  obtained  by 

multiplying  by  3  the  term  which  immediately  precedes  it. 

5.  The  law  may  be  such  that  any  term  may  be  found  when  its 
location  in  the  sequence  is  specified. 

E.  g.    In  the  sequence  1,  4,  9,  16, ,n^ ,  the  seventh  term  is 

72  =  49 ;  the  tenth  term  is  10^  =  100  :  etc. 


620  FIRST  COURSE  IN  ALGEBRA 

III  the  sequence  i,  f ,  f ,  t, ,  -— r  » ,  the  ninth  term  is 

9  9 

'       =  —  ;  the  twentieth  term  is  |^,  etc. 

6.  A  sequence  is  said  to  be  finite  if  it  contains  a  finite  or  limited 
number  of  terms,  and  infinite  if  it  contains  an  infinite  or  unlimited 
number  of  terms. 

I.   Arithmetic  Progression 

7  An  arithmetic  prog^ression  (A.  P.)  is  a  sequence  of  numbers 
each  of  which,  after  the  first,  may  be  obtained  by  adding  to  the 
number  which  precedes  it  in  the  sequence  a  definite  number  called 
the  common  difference. 

E.  g.    In  the  A.  P.  4,  6,  8,  10,  12,  14,  the  common  difference  is  2  ; 
In  —  22,  —  12,  —  2,  8,  18,  the  common  tlifference  is  10  ; 
In  10,  7,  4,  1,-2,  —  5,  —  8,  the  common  difference  is  —  3. 

8.  An  arithmetic  progression  is  said  to  be  increasing  or  de- 
creasing according  as  the  common  difference  is  positive  or  negative. 

9.  In  order  that  the  terms  of  the  sequence,  fl^i,  ^2,  «3,  «*, , 

a„  shall  form  an  arithmetic  progression,  it  is  necessary  that 

ch  —  ai  =  cis  —  02  = =  a„  —  «„  _  1. 

10.  If  (ii  represents  the  first  term,  d  the  common  difference  (that 
is,  the  difference  between  every  two  consecutive  terms),  and  n  the 
number  of  the  place  of  any  specified  term  in  the  progression,  any 
arithmetic  progression  may  be  represented  by  the  general  expression 

ai,  ai  -{-  dj  ai  +  2  dj  ai  -[-  S  dj ,  ai  +  (n  —  l)d. 

11.  The  nth  term.  Observe  that,  since  each  term  of  the  arith- 
metic progression  after  the  first  is  obtained  by  adding  d  to  the  pre- 
ceding term,  the  coefficient  of  d  in  any  specified  term  is  always  less 
by  unity  than  the  number  of  the  term. 

Accordingly,  in  the  nth.  term,  ai  +  (n  —  1)  d,  the  coefficient  of 
d  is  n  —  1, 

Representing  the  nth  term  by  a„,  we  have  the  formula 

Ex.  1.  Find  the  ninth  term,  Og,  of  an  A.  P.  the  first  term  of  which  is 
5  and  common  difference  3. 


ARITHMETIC   PROGRESSION  621 

Substituting  5,  9,  and  3  for  ai,  ?i,  and  d  respectively,  in  the  formula 
above,  we  may  write:  ctg  =  5  +  (9  —  1)  3  =  29. 

Ex.  2.    Find  the  tenth  term  of  the  A.  P.  11,  7,  3,  -  1, 

We  have  a^  =  11,  ?i  =  10,  d  =  —  4. 

Therefore,  a^o  =  11  +  (10  -  1)(-  4)  =  -  25. 

12.  Since  the  formula  <a!„  =  ai  +  (n  —  \)d  contains  four  quantities, 
a„,  «!,  n,  and  d,  it  follows  that  if  any  three  of  them  are  known  the 
fourth  may  be  found. 

Ex.  3.  Write  the  A.  P.  the  first  and  fourteenth  terms  of  which  are  9  and 
87  respectively. 

We  have  a^  =  9,     a{^  =  87,     7i  =  14. 

Hence,  substituting  in  a„  =  aj  +  (u  —  l)tZ,  we  have  87  =  9  +  I3d. 
Hence  d  =  6. 

Hence,  the  required  A.  P.  is  9,  15,  21,  27, ,  81,  87. 

13.  From  the  principles  relating  to  the  solution  of  simultaneous 
linear  equations,  it  follows  that,  if  the  values  of  any  two  of  the 
quantites  «„,  (ti,  w,  or  d  be  unknown,  two  conditional  equations 
are  necessary  to  determine  their  values. 

Hence,  an  arithmetic  progression  may  be  written  if  any  two  of  its 
terms  are  given. 

Ex.  4.  Write  the  A.  P.  the  fourth  and  thirteenth  terras  of  which  are 
29  and  92  respectively. 

We  have  a^  =  29,  and  Ojg  =  92.  Hence,  when  n  =  1,  the  terra  a„ 
represents  29  ;  and  when  n  =  13,  tiie  term  «„  represents  92, 

Substituting  these  values  in  the  formula    a„  =  «i  +  (w  —  1  )<i, 
we  have  29  =  flj  +  (  4  -  l)d    (1), 

and  also  92  =  a^  +  (13  -  l)d     (2). 

Solving  (1)  and  (2),  we  find  that  a^  —  8  and  d  =  l. 

Accordingly  the  required  A.  P.  is  8,  15,  22,  29,  36, ,  85,  92. 

The  required  arithmetic  progression  may  also  l)e  obtained  by  the  follow- 
ing method : 

Since  the  thirteenth  term  of  the  progression  is  the  tenth  term  of  the  pro- 
gression of  which  the  given  fourth  term,  29,  is  the  first  term,  we  may 
reduce  our  problem  to  that  of  finding  an  A.  P.  the  first  term  of  which  is 
29  and  the  tenth  term  of  which  is  92.  After  determining  this  progression 
we  may  write  the  three  necessary  terms  preceding  the  term  29,  now 
considered  as  a  first  term. 

We  have     92  =  29  +  9  d,     hence,     d=l. 


622  FIRST  COURSE   IN  ALGEBRA 

By  writing  three  terms  before  the  "first  term,"  29,  and  nine  terms  after 
29,  we  obtain  the  same  sequence  as  above. 

Exercise  XXVIL     1 

1.  Find  the  12th  and  loth  terms  of  2,  6,  10, 

2.  Find  the  lOth  and  IGth  terms  of  3,  9,  15, 

3.  Find  the  17th  and  11th  terms  of  —  8,  —  3,  2, 

4.  Find  the  7th  and  13th  terms  of  2/3,  1,  4/3, 

5.  Find  the  20th  and  40th  terms  of  2  / 15,  —  1  /  30,  —  1/5, 

6.  Find  the  12th  and  21st  terms  of  —  4,  —  13,  —  22, 

7.  Find  the  10th  and  37th  terms  of  h  h  }h • 

8.  Find  the  8th  and  13th  terms  of  ^-^^,  ^^  '^^^, 

a  a  a 

9.  Find  the  1 4th  and  1 9th  terms  of  (a  —  3),  4  a,  (7  «  +  3), 

10.  Find  the  17th  and  31st  terms  of  (5  a;  —  4),  (2x  —  2),  —  x, 
(-4a: +2), ^ 

Find  the  last  term  in  each  of  the  following  arithmetic  progressions: 

11.  4,  8,  12, to  36  terms. 

12.  8,  2,  —  4, to  93  terms. 

13.  -  23,  -17,-11, to  100  terms. 

14.  1,  1.3,  1.6, to  21  terms. 

15.  (a  +  5  b),  (a  +  3  ^),  («  +  6), to  13  terms. 

Arithmetic  Means 

14.  If  three  numbers  are  in  arithmetic  progression,  the  one  that 
lies  between  the  other  two  is  called  the  arithmetic  mean  of  these 
two. 

If  a,  A,  b,  be  an  arithmetic  progression,  A  is  called  the  arithmetic 
mean  of  a  and  b. 

From  the  A.  P.  represented  by  a.  A,  b,  we  have  by  definition 
A-a  =  b-A. 

Solving  for  A,  we  have  A  =  — - — 

That  is,  ths  arithmetic  mean  of  two  numbers  is  one-half  their  sum. 
E.  g.    The  arithmetic  mean  of  5  and  17  is  11. 

15.  All  of  the  terms  of  an  A.  P.  which  lie  between  two  specified 
terms  are  called  the  arithmetic  means  of  these  two  terms. 


ARITHMETIC  PROGRESSION  623 

E.  g.  Ill  the  A.  P.  2,  3,  4,  5,  6,  7,  8,  the  five  numbers  3,  4,  5,  6,  and  7 
are  called  the  arithmetic  means  of  2  and  8. 

In  the  A.  P.  2,  3^,  4f ,  5|,  6|,  8,  the  four  numbers  3^,  4f,  5f ,  and  6f  are 
called  the  arithmetic  means  of  2  and  8. 

In  the  A.  P.  2,  4,  6,  8,  the  two  numbers  4  and  6  are  called  the  arithmetic 
means  of  2  and  8. 

In  the  A.  P.  2,  5,  8,  the  single  number  5  is  called  the  arithmetic  mean 
of  2  and  8. 

Ex.  1.    Insert  11  arithmetic  means  between  20  and  116. 

The  given  numbers  20  and  116,  taken  together  with  11  arithmetic  means, 
form  an  A.  P.  containing  13  terms. 

Hence,  using  the  formula  a„  =  ai  +  (n  —  l)d,  by  substituting  the  values 
«i3  =  1 16,  ttj  =  20,  and  n  =  13,  we  obtain  1 16  =  20  +  12  d.     Hence  d  =  8. 

Accordingly  the  required  arithmetic  means  are 

28,  36,  44,  52,  60,  68,  76,  84,  92,  100,  108. 

16.  The  student  should  note  the  distinction  between  the  arith- 
metic mean  or  average  of  n  numbers,  and  the  n  arithmetic  means 
inserted  between  two  numbers. 

E.  g.  The  arithmetic  mean  or  average  of  the  five  numbers  2,  6,  15,  50, 
67  is  (2  +  6  +  15  -}-  50  +  67)/5  =  28,  but  these  numbers  do  not  form  an 
A.  P.,  and  cannot  be  regarded  as  being  arithmetic  means  of  any  two 
numbers. 

Exercise  XXVII.    2 

Find  the  arithmetic  mean  of 

1.  26  and  32. 

2.  17  and  11. 

3.  8  and  47.  w.       .      ,   , 

a  +  b 

7.  Find  the  thirty  arithmetic  means  of  18  and  142. 

8-  Find  the  nineteen  arithmetic  means  of—  27  and  113. 

9.  Find  the  seventeen  arithmetic  means  of  25  and  115. 

10.  Find  the  eight  arithmetic  means  of  19  and  23. 

11.  Find  the  twelve  arithmetic  means  of  2  and  3. 

12.  Find  the  fifteen  arithmetic  means  of  2  and  9. 

17.  A  series  is  the  sum  (or  the  limit  of  the  sum)  of  a  succession 
of  numbers,  each  formed  according  to  some  common  law. 

The  successive  numbers  are  called  the  terms  of  the  series. 


624  FIRST  COURSE  IN   ALGEBRA 

E.  g.  If  2,  4,  6,  8, ,  2w  be  a  given  sequence  of  numbers,  the  ex- 
pression obtained  by  writing  the  sum  of  the  successive  terms  of  the  sequence! 
is  called  the  series  2  +  4  +  6  +  8  + +  2w. 

18.  A  series  is  said  to  be  finite  or  iufinite  according  as  the 
number  of  its  tferms  is  finite  or  infinite. 

19.  Although  a  succession  of  numbers  forming  a  sequence  might 
be  described  as  a  series  of  numbers,  the  word  "series,"  as  used  in 
mathematics,  has  especial  reference  to  addition. 

Accordingly,  we  speak  of  the  succession  or  sequence  of  numbers 

«!,  a2>  «8,  «4, y  dn)  but  by  writing  the  sum  of  these  numbers 

we  obtain  the  series  «i  -f  ^^2  +  ^3  +  «4  + +  ctn- 

20.  An  arithmetic  series  is  a  series  the  terms  of  which  are  in 
arithmetic  progression. 

Sum  of  the  Terms  of  an  Arithmetic  Progression 

21.  The  sum  *y„  of  the  terms  of  an  arithmetic  progression  of  n 
terms  may  be  found  as  follows  : 

Sn=  fli  +  (ai  +  fiO  +  («i  +  2^+ +K  -  2 fi?)  +  (««  -  0+<2„, 

>S;=  «,  +  («,  -  tQ  -f  (a»  -  2  (/)  + +  (ai  +  2  qT)  +  («i  +  flQ  +«! . 

By  Addition 

2  AS;=(ai+tf„) +  («!+«„) +  («!+«„)+ +(«l+«„)  +  («l+«»)  +  («l+«n) 

=  w(ai  +  a^. 
Hence 

Sn  =  |(«i  +  any  (1) 

Expressing  the  last  term,  «„,  in  terms  of  ^i,  n,  and  d,  by  means  of 
the  formula  a„  =  «i  +  (n  —  l)d,  the  expression  above  may  be 
written 

Sn  =  ^[«i  +  ^1  +  (n  -  l)d]. 

Or,     Sn  =  '^[2ai  +  (n-l)d].  (2) 

Ex.  1.    Find  the  sum  of  46  terms  of  the  A.  P.  11,  18,  25, 

Substituting  the.  values  n  =  46,  a^  =  II,  and  c?  =  7,  in  the  formula 


ARITHMETIC   PROGRESSION  625 

we  obtain  S^^  =  ^[2  •  11  +  45  •  7]  =  V751. 

Ex.  2.    Find  the  sum  of  the  terms  of  the  arithmetic  series 

5  +  8  +  11  + +95. 

We  have  a^  =  5,  d  =  3,  and  On  =  95. 

To  use  either  of  the  formulas  for  S„,  it  is  necessary  to  know  the  value  of  w, 
which  may  be  found  by  means  of  the  formula  a„  =  rtj  +  (w  —  l)d. 
By  substitution,  95  =  5  +  (>i  —  1)3.     Hence  n  =  3l. 

71 

Substituting  in  the  formula,    <S'„  =  -(a^  +  «„), 
we  have  S^^  =  ^^(5  +  95)  =  1560. 

22.  The  five  numbers  represented  by  a„,  «i,  d,  n,  and  >S'„  are  called 
the  elements  of  an  arithmetic  progrression. 

Ex.3.  Write  the  A.  P.,  having  given  the  elements  /Sj^  =  510  and 
«io  =  87. 

Substituting  510  for  /S,„  87  for  a„  and  10  for  n  in  Sn  =  -x  (ai  +  a„),  we 
obtain,  510  =  5  (aj  +  87).     Hence  a^  =  15. 

To  write  the  required  A.  P.  it  is  necessary  to  know  d.  Using  the 
formula  a„  =  a^  +  (w  —  l)dj  we  find  that  d  =  &. 

Hence  the  required  A.  P.  is  15,  23,  31,  39, ,  87. 

Ex.  4.  How  many  terms  of  the  arithmetic  series  —  16  —  12  —  8  —  4 
+  0  +  4  + must  Ixi  taken  to  obtain  the  sum  72  ? 

Substituting  72  for  ;S„,  —  16  for  a^,  and  4  for  c?  in  /S„  =  -  [2  aj  +  (w  —  l)d]y 
we  obtain  72  =  ^[-  32  +  (n  -  1)4]. 

Or,  72  =  -16?i  +  2w2-2w. 

Solving  this  quadratic  equation  for  w,  we  obtain  n  =  12,  and  n  =  —  3. 

It  will  be  found  that  the  sum  of  the  following  twelve  terms  is  72 : 

-  16,  -  12,  -8,-4,  0,  4,  8,  12,  16,  20,  24,  28. 

It  may  be  observed  that  if,  beginning  with  the  last  term  28,  we  count 
backward  three  terms,  the  sum  is  72. 

We  have  thus  an  interpretation  for  the  negative  value  of  n. 

Ex.  5.    How  many  terms  of  the  A.  P.  39,  34,  29, ,  must  be  taken 

in  order  that  the  sum  shall  be  168  ? 

Substituting  the  values  a^  =  39,  (i  =  —  5,  and  Sn  =  168,  in  the  formula 

,S'^  =  o  [2  «i  +  (»*  —  I)^]»  we  obtain  336  =  78n  —  5n^  -i-  6n,  the  solutions 

of  which  are  found  to  be  w  =  7,  and  n  =  9f . 

40 


626  FIRST  COURSE  IN  ALGEBRA 

It  will  be  found  that  the  sum  of  seven  terms  of  the  given  arithmetic 
progression  is  168. 

The  fractional  value  n  =  9f  may  be  interpreted  as  meaning  that  the  sum 
168  is  greater  than  the  sum  of  nine  terms  and  less  than  the  sum  of  ten 
terms  of  the  series. 

In  this  particular  arithmetic  progression  it  will  be  found  that  168  is  the 
sum  of  nine  terms  increased  by  3/5  of  the  common  difference. 


Exercise  XXVII.     3 

Find  the  sum  of  the  terms  of  each  of  the  following  arithmetic 
progressions  : 

1.  3,  8,  13, ,33.  5.  i  f ,  1, ,  39f  • 

2.  20,  18,  16, ,0.  6.  I,  0,  -  ^, ,  -  24^. 

3.  25,  23,  21, ,  -  15.  7.  2/3,  14/15,  6/5, ,6. 

4.  19,  32,  45, ,  188.  8.  7/12,  7/6,  7/4, • ,  7. 

9.  6,  10,  14, to  31  terms. 

10.  -  23,  —  27,-31, to  19  tenns. 

11.  —  17,  —  6,  5, to  13  terms. 

12.  —  32,  23,  78, to  15  terms. 

13.  7/2,  9/2,  11/2, to  16  terms. 

14.  1/2,  0,  -  1/2, to  25  terms. 

15.  3,  4^,  6, to  83  terms. 

16.     J     >     >     to  a  terms. 

a  a  a 

11.    (c  -\-  d),  {-  c  +  2  d),  (-  3c  -\-  Sd),     to  60  terms. 

18.  Find  the  sum  of  all  of  the  even  numbers  from  20  to  80  in- 
clusive. 

19.  If  ai  =  22,  d=2,  and  >S;  =  820,  find  n. 

20.  Find  c?,  knowing  that  «i  =  9  and  an  =  29. 

21.  If  >Su  =  66  and  «„  =  23,  find  ««. 

22.  If  r(i  =  16  and  d  =  —  5,  find  the  terms  the  values  of  which 
lie  between  —  70  and  —100. 

23.  Find  «2o,  having  given  «6  =  2  and  a^  =  —  2. 

24.  Find  the  sum  of  the  terms  the  values  of  which  lie  between  0 
and  50  of  the  series  the  fourth  term  of  which  is  13  and  the  common 
difference  of  which  is  —  7. 


HARMONIC   PROGRESSION  627 

25.  The  product  of  three  numbers  in  arithmetic  progression  is  48, 
and  the  first  number  is  three  times  the  last.     Find  the  numbers. 

26.  Of  three  numbers  which  are  in  arithmetic  progression  the 
third  is  eleven  times  the  first.  Find  the  numbers  if  the  sum  of  the 
three  numbers  is  equal  to  the  eighteenth  term  of  the  arithmetic 
progression  —  18,  —  16,  —  14, 

27.  The  sum  of  three  numbers  in  arithmetic  progression  is  30 
and  the  sum  of  their  squares  is  350.     Find  the  numbers. 

28.  Find  the  number  of  terms   in  the  arithmetic  progression 

1,  9,  17, ,  the  sum  of  which  approximates  most  closely  to 

1000. 

29.  Find  the  sum  of  all  of  the  multiples  of  7  which  lie  between 
zero  and  200. 

30.  Show  that  the  sum  of  the  first  n  odd  numbers  is  equal  to  n\ 

Problems  in  Physics 

31.  A  car  starting  from  a  state  of  rest  moves  down  an  inclined  track, 
passing  over  distances  of  1  foot  the  first  second,  3  feet  the  second  second, 
5  feet  the  third  second,  etc.     Find  the  distance  passed  over  in  one  minute. 

32.  A  ball  starting  from  a  state  of  rest  rolls  down  an  inclined  hoard, 
passing  over  distances  of  5  inches,  15  inches,  25  inches,  etc.,  in  successive 
seconds.  Find  the  number  of  seconds  required  for  the  ball  to  pass  over 
a  distance  of  15  feet. 

33.  If  a  ball,  starting  up  an  inclined  plane,  passes  over  40  feet  the  first 
second,  36  feet  the  second  secoiul,  32  feet  the  third  second,  etc.,  find  the 
number  of  seconds  required  by  the  ball  to  pass  over  a  distance  of  196  feet. 

34.  It  is  found  that  when  a  ball  is  thrown  vertically  upward  the  force  of 
gravity  diminislies  the  distance  passed  over  in  successive  seconds  by  32  feet 
per  second  (nearly).  Find  the  distance  passed  over  in  4  seconds  by  a  ball 
which,  when  thrown  vertically  upward,  rises  to  a  height  of  128  feet  during 
the  first  second. 

35.  If  the  force  of  gravity  increases  the  space  passed  over  by  a  faUing 
body  in  successive  seconds  by  32  feet  per  second,  find  the  distance  passed 
over  in  6  seconds  by  a  falling  body  which,  when  thrown' downward,  passes 
over  a  distance  of  24  feet  during  the  first  second. 

11.     Harmonic  Progression 

23.  A  liarmonic  progression  (IT.  P.)  is  a  sequence  of  numbers 
the  reciprocals  of  which  are  in  arithmetic  progression. 


628  FIRST  COURSE  IN  ALGEBRA 

A  harmonic  series  is  a  series  of  numbers  the  terms  of  which 
are  in  harmonic  progression. 

Principle :  If  three  numbers,  represented  by  «,  ft,  and  c,  are  in 
harmonic  progression,  it  follows  that  a  :  c  =  a  —  b  :  b  —  c.     (l) 

For,  if  a,  b,  c  are  in  harmonic  progression,  it  follows  by  definition 

that  -,  T,  -,  is  an  arithmetic  progression. 
a    0    c 

Hence,  we  have 


1 
b~ 

1 
a 

_  1 
c 

1 
b 

a  - 

-b 

_b- 

-  c 

Hence,  ,     —     , 

ab  be 

Or  c{a  —  b)  =  a(b  —  c). 

That  is,  a:  c  =  a  —  b  :b  —  c, 

24.  The  numbers  of  any  sequence  are  in  harmonic  progression 
if  every  three  consecutive  numbers  are  in  harmonic  progression. 

25.  When  three  numbers  form  a  harmonic  progression  the 
middle  number  is  called  the  harmonic  mean  of  the  other  two. 

E.  g.    Th(i  harmonic  mean  of  1/2  and  1/4  is  1/3. 

26.  The  harmonic  mean  of  two  numbers,  represented  by  a  and  6, 
may  be  found  as  follows  : 

Representing  the  harmonic  mean  by  H,  we  have  the  harmonic 
progression,  a,  H,  b. 

Accordingly,  -,  7^,  7,  must  be  an  arithmetic  progression. 
a    H    h 

Hence  H-a^b~H' 

Solving  for  H,  we  obtain  H  =  • 

Ex.  1.    Find  the  harmonic  mean  of  4  and  12. 

Substituting  4  for  a  and  12  for  h  in  the  formula  H  =  2  ab  /  (a  +  h),  we 
obtain  6. 

27.  In  any  harmonic  progression  all  of  the  terms  lying  between 
any  two  specified  terms  are  called  the  harmonic  means  of  these 
two  terms. 

E.  g.    3/7,  3/8,  1/3,  3/10,  3/11  are  harmonic  means  of  1/2  and  1/4. 


HARMONIC   PROGRESSION  629 

28.  Problems  in  harmonic  progression  are  generally  solved  by- 
obtaining  the  reciprocals  of  the  terms  and  making  use  of  the  proper- 
ties of  the  resulting  arithmetic  progression. 

Ex.  2.  Find  the  12th  term  of  the  harmonic  progression  1/4,  1/7,  1/10, 
1/13, The  reciprocals  of  the  terms  of  the  given  harmonic  pro- 
gression form  the  arithmetic  progression,  4,  7,  10,  13, ,  the  12th  term 

of  which  is  found  to  be  37. 

Accordingly,  the  required  term  of  the  given  harmonic  progression  is  1/37. 

29.  There  is  no  general  formula  for  the  sum  of  the  terms  of  a 
harmonic  progression. 

Exercise  XXVII.    4 

1.  Find  the  8th  term  of  1/2,  1/3,  1/4, 

2.  Find  the  6th  term  of  1/50,  1/65,  1/80, 

8.    Find  the  17th  term  of  2,  3/2,  6/5, 

4.  Find  the  4th  term  of  the  H.  P.  the  first  term  of  which  is  1/51 
and  the  13th  term  of  which  is  1/3. 

5.  Find  the  17th  and  18th  terms  of  the  H.  P.  the  second  and 
sixth  terms  of  which  are  1/11  and  1/27  respectively. 

6.  Find  the  H.  P.  in  which  the  6th  term  is  1/7  and  the  11th 
term  1/13. 

7.  Find  the  H.  P.  in  which  the  37th  term  is  1/74  and  the  13th 
term  is  1/26. 

8.  Find  the  H.  P.  in  which  the  9th  term  is  1/4  and  the  15th  term 
is  —  1/14. 

9.  Find  the  H.  P.   in  which  the  third   term   is  5/6  and  the 
sixth  term  is  1/3. 

10.  The  first  two  terms  in  a  harmonic  progression  are  14  and  7. 
Find  the  number  of  terms  which  lie  between  —  7  and  —  2. 

11.  Find  the  harmonic  mean  of  3  and  9. 

12.  Find  the  harmonic  mean  of  1/7  and  1/8. 

13.  Find  the  harmonic  mean  of  1/20  and  1/30. 

14.  Find  the  two  harmonic  means  of  3  and  10. 

15.  Insert  4  harmonic  means  between  5  and  15. 

16.  Insert  3  harmonic  means  between  1/4  and  1/324. 

17.  Insert  5  harmonic  means  between  1/7  and  1/22. 

18.  Insert  7  harmonic  means  between  1/9  and  1/65. 


630  FIRST  COURSE   IN  ALGEBRA 

19.  Find  two  numbers  the  sum  of  which  is  20  and  the  harmonic 
mean  of  which  is  15/2. 

20.  The  difference  between  the  arithmetic  and  harmonic  means 
of  two  numbers  is  99/10,  and  one  of  the  numbers  is  four  times  the 
other.     Find  the  numbers. 

21.  If  X  -{■  1/,  y  -\-  z  and  z  +  x  form  a  harmonic  progression,  show 
that  i/'\  x^  and  z^  form  an  arithmetic  progression. 

III.    Geometric  Progression 

30.  A  geometric  progression  (G.  P.)  is  a  sequence  of  numbers 
each  of  which,  after  the  first,  may  be  obtained  by  multiplying  the 
number  which  precedes  it  in  the  sequence  by  some  particular 
multiplier. 

31.  From  this  definition  it  follows  that  the  quotient  obtained  by 
dividing  any  term  in  the  progression,  after  the  first,  by  the  one 
which  immediately  precedes  it,  is  the  same  for  every  two  consecu- 
tive terms.  The  quotient  thus  obtained  is  called  the  common 
ratio  and  is  usually  denoted  by  ?\ 

E.  g.    In  the  G.  P.  2,  4,  8,  16,  32,  the  common  ratio  is  2. 

In  81,  27,  9,  3,  1,  1  /3,  1/9,  the  common  ratio  is  1/3. 

In  100,  -  20,  4,  -  4  /5,  4/  25,  -  4/125,  the  common  ratio  is  -  1  /5. 

32.  If  rtTi  represents  the  first  term,  r  the  common  ratio,  and  n  the 
number  of  the  place  of  any  specified  term  in  the  progression,  any 
geometric  progression  may  be  represented  by  the  general  expression 
«i,  (fir,  air\  «l^^  air\ ,  air"'^. 

33.  Observe  that  in  any  specified  term  the  exponent  of  the  power 
to  which  r  is  raised  is  less  by  unity  than  the  number  of  the  term. 

E.  g.    r^  appears  in  the  fourth  term ;  r®  in  the  tenth  term,  etc. 

34.  The  wtli  term.  It  appears  that  a  particular  term,  repre- 
sented by  «„,  may  be  calculated  by  means  of  the  formula 

That  is,  in  any  geometric  j^rogresdon  any  term  can  be  found  by 
multiplying  the  first  term  by  the  common  ratio  raised  to  a  power  the 
exponent  of  ujhich  is  equal  to  a  number  which  is  less  by  unity  than  the 
number  of  the  required  term. 


GEOMETRIC   PROGRESSION  631 

Ex.  1.    Find  the  eighth  term  of  the  geometric  progression  3, 6, 12, 

The  common  ratio  is  obtained  by  dividing  any  one  of  the  terms  by  the 
term  immediately  preceding  it,  for  example  6-^3  =  2. 

Substituting  8  for  w,  3  for  a^,  and  2  for  r  in  the  formula  a„  =  «ir«~i, 

we  obtain  ag  =  3  •  2'  =  384. 

Ex.  2.    Find  a^,  having  given  a^  =  12288  and  a^  =  768. 
Using  the  formula  a„  =  «i?  «~i  we  obtain  two  conditional  equations  in 
which  ttj  and  r  may  be  regarded  as  unknowns. 

12288  =  air«,  (1)  and  768  =  a^r*.     (2) 

Dividing  the  members  of  equation  (1)  by  the  corresponding  members  of 
equation  (2),  we  obtain  16  =  r^.     Hence  r  =  ±  4. 
Substituting  for  r  in  (2),  we  obtain  a^  =  3. 
Substituting  in  rtg  =  AjT,  we  obtain  the  required  second  term. 

a2  =  S  (±4)  =  ±  12. 

35.  A  geometric  progression  is  said  to  be  finite  or  infinite 
according  as  the  number  of  its  terms  is  finite  or  infinite. 

E.  g.    1,  3,  9,  27,  81,  is  a  finite  G.  P.  the  common  ratio  of  which  is  3. 

1,  3,  9,  27,  81, ,  is  an  infinite  G.  P.  the  common  ratio  of  which  is  3. 

81,  27,  9,  3,  1,  1/3,  1  /9,  1/27, ,  is  an  infinite  G.  P.  the  common 

ratio  of  which  is  1  /3. 

36.  A  geometric  progression  is  said  to  be  increasing  or  de- 

creasinjf  according  as  its  successive  terms  increase  or  decrease. 

The  successive  terms  of  a  geometric  progression  increase  or  de- 
crease according  as  the  common  ratio  is  greater  than  or  less  than 
unity. 

37.  Representing  the  first  term  of  a  geometric  progression  by  «i 
and  the  common  ratio  by  r  it  may  be  seen  that  the  following  general 
expression  may  be  used  to  represent  any  geometric  progression  : 

The  values  of  the  different  terms  depend  entirely  upon  the  values 
which  may  be  assigned  to  the  letters  a^  and  r,  considered  as  inde- 
pendent variables. 

Hence,  in  general,  a  geometric  progression  is  completely  deter- 
mined when  two  independent  conditions  affecting  the  values  of  its 
terms  are  given. 


632  FIRST  COURSE   IN  ALGEBRA 

Exercise  XXVIL     5 

Find  the  terms  specified  in  each  of  the  following  geometric 
progressions  : 

1.  The  8th  and  9th  terms  of  2,  4,  8,  16, 

2.  The  9th  and  11th  terms  of  20,  10,  5, 

3.  The  6th  and  10th  terms  of  1/2,  1/4,  1/8, 

4.  The  6th  and  12th  terms  of  1,  1/3,  1/9, 

5.  The  5th  and  13th  terms  of  3,  —  6,  12, 

6.  The  6th  and  14th  terms  of  100,  50,  25, 

7.  The  10th  and  20th  terms  of  .1,  .01,  .001, 

8.  The  15th  and  30th  terms  of  m,  m%  m\ 

9.  The  12th  and  40th  terms  of  1/a*,  l/«*,  l/a^ 

10.  The  19th  and  51st  terms  of  1,  1/c,  l/c^ 

11.  The  11th  and  14th  terms  of  a;,  —  i/y  y^jx, 

12.  The  cth  and  6th  terms  oiajb,  ajbc,  a/bc^ 

Geometric  Means 

38.  When  three  numbers  are  in  geometric  progression,  the  middle 
number  is  called  the  geometric  mean  of  the  other  two. 

E.  g.  In  the  geometric  progression  2,  4,  8,  the  geometric  mean  of  2  and 
8  is  4;  in  the  geometric  progression  1,-6,  36,  the  geometric  mean  of 
1  and  36  is  —  6. 

39.  The  geometric  mean  of  any  two  numbers,  a  and  6,  may  be 
found  as  follows  : 

Denoting  the  geometric  mean  of  a  and  b  by  G,  we  have  the  geometric 
progression  a,  Gy  b. 

By  the  definition  of  a  G.  P.  we  have  G  -^  a  =  b  -^  G. 

Hence,  G  =  ±  \/ah. 

That  is,  the  geometric  mean  of  two  numbers  is  the  sqaare  root  of 
their  product. 

It  should  be  observed  that,  since  the  common  ratio  of  a  geometric  progres- 
sion may  be  either  positive  or  negative,  the  successive  terms  of  a  geometric 
progression  may  be  either  all  positive  or  all  negative,  or  alternately  positive 
and  negative.  Accordingly,  the  double  sign  ±  should  be  employed  before  the 
radical  sign  in  the  expression  for  the  geometric  mean  G^  =  ±  ^ab, 

(Compare  with  Chapter  XXVI.  §  24  (ii.).) 


GEOMETRIC  PROGRESSION  633 


Ex.  1.    Find  the  jjeoraetric  mean  of  9  and  16. 


Using  the  formula  G  =  ±  ^ab,  we  have  G  =  ±  \/9  •  16  =  ±  12. 

40.  In  any  geometric  progression  all  of  the  terms  lying  between 
any  two  specified  terms  are  called  the  geometric  means  of  these 
two  terms. 

Ex.  2.  Find  the  four  geometric  means  of  1/32  and  32. 

Taken  together  with  the  four  geometric  means,  1/32  and  32  may  be 
regarded  as  the  first  and  sixth  terms,  respectively,  of  a  geometric  progression. 

Substituting  1/32  for  a^  and  32  for  a^  in  the  formula  a,t  =  a^r^'^ 
we  have  •  32  =  ^^  r^ 

or  r«  =  32  •  32 

Hence  r  =  4. 

Writing  the  geometric  progression  the  first  term  of  which  is  1/32  and 
c(mimon  ratio  4,  we  obtain  1/32,  1/8,  1/2,  2,  8,  32. 

Accordingly,  the  required  geometric  means  are  1/8,  1/2,  2  and  8. 

41.  Representing  the  arithmetic  mean  of  two  numbers,  a  and  b,  by 
Af  the  geometric  mean  by  (?,  and  the  harmonic  mean  by  H^  we  have 

^=^*.  (1) 

O  =  Va6,  (2) 

^=^.  (3) 

From  (1)  and  (2),  A-G  =  ^-^-^  -  v^  (4) 

it 


a  —  2-v/a&  +  h 
2 


(5) 

(6) 


The  expression  {^/a  —  a/^)^/2  is  positive  if  a  and  b  are  real  num- 
bers and  a  ^  b. 

It  follows  that  A  is  greater  than  G  for  real  values  of  a  and  h 
which  are  such  that  a  i^  b. 


54                        FIRST  COURSE   IN  ALGEBRA 

From  (1)  and  (3), 

2        a  +  6 

From  (2), 

From  (9)  and  (8), 

G'  =  ah. 
G'  =  A'ff. 

Hence, 

A       G 
G~  H 

(7) 

(8) 

(9) 
(10) 

(11) 

That  is,  the  geometric  mexin  of  any  two  numbers  is  also  the  geometric 
mean  of  the  arithmetic  and  harmonic  mean^  of  the  same  numbers. 
Since  A>G\t  follows,  from  (11),  that  G  >  H. 
Hence^  for  any  positive  numbers^  A  >  G  >  II. 

E.  «;.    If  a  =  1  and  b  -  49,  we  have  ^  =  25,  (?  =  7,  and  H  =  l^. 

It  should  be  observed  that  25  >  7  >  1||. 

Exercise  XXVIL     6 

Find  the  geometric  mean  of : 

1.  4  and  16.  5.  2  and  32. 

2.  4  and  25.  6.  2  and  50. 

3.  100  and  1.  7.  1/2  and  32. 

4.  20  and  5.  8.  1/2  and  1/8. 
9.  Find  the  two  geometric  means  of  —  8  and  64. 

10.  Find  the  two  geometric  means  of  3  and  192. 

11.  Find  the  two  geometric  means  of  1/20  and  25/4. 

12.  Find  three  geometric  means  of  27/8  and  2/3. 

13.  Find  three  geometric  means  of  6  and  486. 

14.  Find  the  four  geometric  means  of  1  and  1024. 

42.  A  geometric  series  is  a  series  of  numbers  the  terms  of 
which  are  in  geometric  progression. 

43.  A  geometric  series  is  said  to  be  increasing  or  decreasing 
according  as  its  terms  form  an  increasing  or  decreasing  geometric 
progression. 

Sum  of  the  Terms  of  a  Geometric  Progression 

44.  We  can  obtain  an  expression  for  the  sum,  Sn,  of  n  terms  of  a 
given  finite  geometric  progression,  as  follows  ; 


GEOMETRIC   PROGRESSION  635 

>Sn  =«i    +rtir  +  air^  +  air^  + +  air"-"  +ai?'"-^  (1) 

rS^  =  ai?-  +  air^  +  ay  + +  ai?-'"'^  +«ir'^~^+^ir'*  (2) 

Hence,  rSn  —  S„  =  ay  —  ai.  (3) 

Or,  S,(r  -  1)  =  aiCr**  -  1).  (4) 

Therefore,  ^^^^ai(rH-l) 

r  —  1  ^  ^ 

It  should  be  observed  that  (2)  is  obtained  by  multiplying  both 
members  of  (1)  by  r,  and  (8)  results  from  subtracting  the  members 
of  (1)  from  the  corresponding  members  of  (2). 

46.  When  the  first  term  ai  and  the  wth  term  a,,  are  given,  it  is 
convenient  to  employ  an  alternative  form  for  the  sum  S„. 

This  may  be  obtained  as  follows  : 

The  second  member  of  the  equation    S^  =  may  be 

expressed  in  the  form  — — ^  • 

r  —  1 

That  is,  ^.^^^ir"-^i^ 

r  —  1 
The  first  term  ay  of  the  numerator  may  be  transformed  as 
follows  :  ay  =  ray~^  =  a,{r. 

Hence,  we  have,  8n  =    "^  _      •  (7) 

46.  The  sum  of  a  finite  number  of  terms  of  an  arithmetic  pro- 
gression or  of  a  geometric  progression  may  be  obtained  by  adding 
the  consecutive  terms  as  written. 

When  a  formula  is  used  it  is  unnecessary  to  add  the  terms 
separately.     Hence,  a  formula  is  a  labor-saving  device. 

Ex.  1.    Find   the   sura   of    7   terms   of    the    geometric    progression    5, 

15,  45, 

a^Or^  —  1) 
Substituting  5  for  dj,  3  for  r  and  7  for  n  in  the  formula  >S„  = r-\' 

we  have  ^^  =  ^^'"^^  ~  ^  =  5465. 

47.  If  r  =  1,  all  the  terms  of  a  given  geometric  progression  are 
equal,  and  hence  the  sum  of  n  terms  is  w^i. 


636  FIRST  COURSE  IN  ALGEBRA 

By  taking  n  great  enough,  the  sum  noi  can  be  made  greater  than 
any  assignable  number ;  hence  it  can  be  made  infinitely  great. 

48.  By  taking  n  great  enough,  the  sum  of  n  terms  of  a  geometric 
pi'ogression  in  which  r  is  numerically  greater  than  unity  may  he 
made  to  exceed  any  assrignable  positive  number. 

For,  it  appears  from  S,  =  "^^^"^  ~  ^  =  ^  -  -^  that  if  r 

has  a  value  which  is  numerically  greater  than  unity,  the  value  of 
7-",  and  accordingly  of  air^/{r  —  l),  may  be  made  as  great  as  we 
please  by  taking  the  value  of  w  great  enough. 

Hence  the  value  of  S„  may  be  made  greater  than  any  assignable 
number,  for  r  >  1. 

49.  The  values  represented  by  the  five  quantities  an,  «i,  r,  n,  and 
S^  are  called  the  elements  of  a  geometric  progression. 

It  may  be  seen  that  w,  representing  the  number  of  terms  of  a 
geometric  progression,  must  be  a  positive  integer,  while  the  remain- 
ing elements,  a„,  ai,  r,  and  S^,  may  be  positive  or  negative,  integral 
or  iractional. 

Exercise  XXVII.     7 

Find  the  sum  of  the  terms  of  each  of  the  following  progressions  : 

1.  2,  4,  8, 512. 

2.  20,  —  10,  5, to  6  terms. 

3.  1/2,  1,  2, to  8  terms. 

4.  1/2,  1/2^  l/2», to  9  terms. 

5.  12,  4,  4/3, to  7  terms. 

6.  1/5,  1/5^  l/5«, to  10  terms. 

7.  224,  -  168,  126, to  5  terms. 

8.  6,  —  4,  8/3, to  1 1  terms. 

9.  15,  —  1/3,  1/15, to  6  terms. 

10.  2a/3,  6a/6,  36 V3, to  8  terms. 

Sum  of  the  Terms  of  an  Infinite  Decreasing  Geometric 
Progrcfssion 

50.  By  changing  the  signs  in  both  numerator  and  denomi- 
nator of  the  formula  for  the  sum  of  n  terms  of  a  finite  geometric 
progression. 


GEOMETRIC 

PROGRESSION 

'S'„ 

r  — 

-1) 
1     ' 

-s,. 

_      ^1 

a,r- 

63T 


we  obtain 

\  —  r       i  —  r 

We  shall  for  convenience  of  proof  regard  «i  as  being  positive. 

If  7'  has  a  numerical  value  less  than  unity,  the  absolute  value  of 
aii^  and  accordingly  of  ai/'"/(l  —  r)  will  decrease  as  n  increases  in 
value. 

Accordingly,  by  giving  to  n  a  value  great  enough,  we  may  make 
the  value  of  air"/(l  —  r)  as  small  as  we  please,  but  we  can  never  in 
this  way  make  the  value  of  the  fraction  zero. 

As  the  value  of  the  fraction  «!/'"/( 1  —  r)  diminishes,  the  sum  S^ 
approaches  more  nearly  the  value  of  the  first  fraction  axjil  —  r), 
but  it  never  becomes  exactly  equal  to  it,  because  «ir"/(l  —  ?')  can 
never  become  zero. 

We  may,  by  taking  n  great  enough,  make  the  sum  become  and 
remain  as  nearly  equal  to  ai/(l  —  r)  as  we  please. 

This  is  expressed  by  saying  "  the  limit  of  the  sum  of  an  infinite 
number  of  terms  of  a  decreasing  geometric  progression  is  «i/(l  —  r)." 

Expressed  in  symbols,  we  have:  lim>S^oo  = .  _  .>  which  is  read, 

"  the  limit  of  the  sum  of  an  infinite  number  of  terms  (of  a  given 
geometric  progression)  is  equal  to  «i/(l  —  r)." 

As  an  alternative  form  we  have   Src  —  :; ,  which  is  read, 

1  —  r 

"  the  sum  of  an  infinite  number  of  terms  (of  a  given  geometric  pro- 
gression) approaches  - — —  as  a  limit." 

Ex.  1.     Find  the  sum  of  an  infinite  number  of  terms  of  the  decreasing 

geometric  progression  1,  1/2,  1/4,  1/8, 

Substituting  the  value  1  for  aj,  and  1/2  for  r,  in  the  formula 

/S„  =  T—^ '     we  obtain     S^  =  2. 
^        \  —  r 

Ex.  2.    Find  the  sum  of  an   infinite  number  of  terms  of  36,  —  12,  4, 

-  4/3, 

We  have  a^  =  36,  r  =  -  1/3. 


638  FIRST  COURSE  IN  ALGEBRA 


Hence  from  the  formula,        S^  = 


1  -r 
36 


we  have  ^«  -  1  +  1/3 

Or,  S^  =  27. 

51.  By  means  of  the  formula  for  the  sum  of  an  infinite  number 
of  terms  of  a  decreasing  geometric  progression  we  may  obtain  the 
ffeneratinjf  fractiou  of  a  repeating  decimal  fraction,  that  is, 
the  fraction  which  gives  rise  to  a  repeating  decimal  fraction  if 
the  numerator  is  divided  by  the  denominator. 

Ex.  3.   Find  the  generating  fraction  of  the  repeating  decimal  fraction  .3. 

It  should  be  observed  that  the  dot  written  above  the  3  indicates  that  3  is 
to  be  repeated  indefinitely ;  ,that  ia,  .3  =  .333333  +. 

We  may  write  .33.33 =  ^^4.^^^  +  ^^^  +  ^^^  + 

In  this  form  the  repeating  decimal  fraction  appears  as  a  decreasing  geo- 
metric series  the  first  term  of  which  is  3/10,  and  the  common  ratio  of  which 
is  1/10. 

Using  the  formula  for  the  sum,    S^  =  - — ^ 

we  have    *S^  = 


1  -  1/10 
Or    S^=l 

Ex.  4.  Find  the  improper  fraction  which  may  be  transformed  into  the 
repeating  decimal  fraction  3.236. 

We  may  write  3.236  =  3.2  +  .036. 

It  should  be  understood  that  the  dots  above  the  3  and  6  denote  that  36 
is  to  be  repeated  indefinitely,  that  is,  .036  =  .0363636363636  +  . 

Accordingly,  we  have        .036  =  yff^  +  j-^U^  +  T^lfolTo  + » 

from  which     a^  =  36/1000,     and     r  =  1/1(X). 


Usinsr  the  formula 


«i 


1-r' 


, ,  .  ^      .     36/1000 

weobtam  .^^^____. 

The  required  fraction  may  be  obtained  by  finding  the  sum  of  3.2  and 
2/55,  which  is  found  to  be  178/55. 


GEOMETRIC  PROGRESSIO]^  630 

The  student  should  show,  by  dividing  the  numerator  by  the  denominator, 
that  the  fraction  178/55  gives  rise  to  the  given  repeating  decimal  fraction 
3.23636-f. 

52.  The  process  of  finding  the  generating  fraction  corresponding 
to  any  given  repeating  decimal  fraction  is  sometimes  spoken  of  as 
evaluating  the  given  repeating  decimal  fraction. 

Exercise  XXVII.     8 

Find  the  sum  of  an  infinite  number  of  terms  of  each  of  the  fol- 
lowing series  : 

1.  1/2  +  1/4  +  1/8  + 

2.  1  -  1/2  +  1/4  - 

3.  2/3  +  2/9  +  2/27  + 

4.  500  +  100  +  20  + 

5.  5  -  1/2  +  1/20  -  1/200  + 

6.  2-4  +  8-16  + 

7.  1  —  aj  +  a;^  -  a;»  + ,  for  a;  <  1. 

8.  1-  1/3  +  1/9   - 

9.  1  +  1/a;  +  llx"  + ,  for  a;  >  1. 

Evaluate  the  following : 

10.  .3.  13.  .61.  16.  3.279. 

11.  .6.  14.  .i23.  17.  7.543. 

12.  .25.  15.  .227.  18.  2.564. 

19.  Find  r,  having  given  nfg  =  6  and  a^  =  384. 

20.  Find  two  numbers  the  difference  of  which  is  48  and  the 
geometric  mean  of  which  is  7. 

21.  Find  >Vio,  having  given  «4  =  72  and  a-,  =  —  64/3. 

22.  Find  ai,  provided  that  ««  =  1/32  and  a^  =  —  1/968. 

23.  Find  «9,  knowing  that  a^  =  .008  and  a^  =  .000064. 

24.  The  difference  between  two  numbers  is  70  and  their  arith- 
metic mean  exceeds  their  geometric  mean  by  25.     Find  the  numbers. 

25.  Find  three  numbers  in  geometric  progression  such  that  their 
sum  shall  be  14  and  the  sum  of  their  squares  84. 

26.  The  sum  of  the  first  four  terms  of  a  geometric  progression  is 
15,  and  the  sum  of  the  next  two  terms  is  48.     Find  the  progression. 

27.  A  number  consists  of  three  figures  in  geometric  progression. 


640  FIRST  COURSE  IN  ALGEBRA 

The  sum  of  the  figures  is  7,  and  if  297  be  added  to  the  number  the 
order  of  the  figures  will  be  reversed.     Find  the  number. 

28.  A  ball  is  thrown  vertically  upward  to  a  height  of  120  feet 
and  after  falling  it  rebounds  one-third  of  the  distance,  and  so  on. 
Find  the  whole  distance  passed  over  by  the  ball  before  it  comes 
to  rest. 

Mental  Exercise  XXVII.     9 

Classify  each  of  the  following  as  Arithmetic,  Harmonic,  or 
Geometric  Progressions  : 


1.  1,  4,  7, 28.  1/2,  1/3,  2/9, 

2.  12,  16,  20, 29.  —  1,  1,  8,  •  • 

3.  23,  17,  11, 30.  -1,  1,  1/3,  • 

4.  4,  12,  36, 31.  1,  3,  9,     •  .  . 

5.  5,  20,  80, 32.   1,  1/3,  1/9,  • 

6.-15,-12,-9, 33.  5,  1,  —  3,  •  . 

7.-7,-11,-15, 34.  5,  1,  1/5,  •  •  • 

8.  2,  -  4,  8, 35.  5,  1,  5/9,  •  •  • 

9.  —  3,  6,  —  12, 36.  4,  6,  12, 

10.  2,  5/2,  3, 37.  3,  5,  15, 

11.  2,  2/5,  2/9, 38.  6,  9,  18,  •  •  . 

12.  1/3,  1/6,  1/9, 39.  6,  4,  3, 

13.  3/7,  3/5,  3/2, 40.  20,  15,  12,  •  • 

14.  1/3,  1/9,  1/27, 41.  1,  a;,  ar*, 

15.  1/20,  1/10,  1/5, 42.  a;,  2  a;,  3  a;,  .  • 

16.  1/4,  1/2,  1, 43.  tf^  a\  a\  ■  ■ 

17.  1/4,  1/2,  3/4, U.  b%  h\  6»,  .  .  . 

18.  1/3,  1,  3, 45.  c\  c\  c^\  •  .  . 


19.  1/5,  2,  20, 46.  6/2  a,  b/Aa,  b/Qa, 

20.  15,  3,  3/5, 47.  ^,  c?«,  d^\ 

21.  21,  28,  35, 48.  a^  a%  a^b^,  •  •  •  • 

22.  1/2,  2/5,  1/3, 49.  a,  ab'',  ab\ 


23.  1/2,  2/5,  8/25, 50.  a;/5,  a;/10,  a;/15, 

24.  1/2,  1/3,  1/4, 51.  1,  1/a;,  1/a;^  •  •  • 

25.  1/2,  1/3,  1/6, 52.  1///,  l/6^  llb\ 

26.  2,  3,  6, •  53.  llx\  4/a;^  7/a;^ 

27.  6,  3,  2, '  54.  c,  1,  1/c, 


REVIEW  641 

55.  «/2,  a/S,  al4:, ^^  1  2 

56.  a,  6,  b^ja, b  b 

57.  3M  4/x,  5/a;, ^,    a-l   a~2  «-3 

Pie    ^.-1   iy.-2  ^-3 71.  > }  — , , 

59.  ajbcy  a/c,  ab/c^ 


c         2c      4c 


61.  a,  a/bj  a/h^ 

62.  a,a  +  b,a  +  2b, 73.  ai-b,a^+ab,a^+a% 

QS.  x,x-y,x-2^, ^^^  2 

64.  a  +  2,a-2,a-6, 74.  -y- ,  1'^+"^' 

65,  a;  —  6,  ic  —  3,  £c, 

Q6.  a  —  2b,a  —  b,a, 75.  — £_,  1,  £JZi?, 

67.  a+ 26,  «,  «-26,-  •  •  .  .  .         'c-d      '      c 

68.  ?w,2  7w+w,3m+2;2,-  •  •  •  •  .  „.       6 

69.  a^«^-^.^a^-26^ ^^'  ^^'  ^'  ^^-^^^  "" 

Exercise  XXVII.  10.    Review 
Simplify  each  of  the  following  : 

5.  («  ^  —  6  ^)  -7-  («  —  6).  6. 


'■  (S)ll)~(l) 


a-1  +  b- 

a+b 


8.  If«  =  -1,  ^)  =  -2,  c  =  -3,  fmdthevalueof^  +  —  + -^ 
'  '  '  be      ca      ab 


9.  Simplify  [(a;  +  y) V^~^][(a;  —  2/) V^  +  3^]- 


10.  Simplify  (V^^  _  /)(Va;  +  y)(va;  -  3^)- 

11.  Simplify  (v^  +  Vb){^/a+  v^)(v^a  -  V^). 

41 


642  FIRST  COURSE  IN  ALGEBRA 

12.  Show  that  (a  4-  3)(rt  +  4)(a  +  5)(a  H-  6)  +  1  =  (a'  +  9  a  +  19)' 

13.  Show  that  (a;  +  2)(a;  +  'S)(x  +  4)(x  +  5)  +  1  is  a  square. 

14.  Show  that  V2  w  +  2  ^/m^  —  n^  =  Vm  +  n  +  ^m  —  n. 


15.  Show  that 

16.  Show  that 


^/a  —  ^/h        ^/a  —  b 

?/  n-f-l/—  n/—   n+2/— 

5  +  3a/5 


17.  Rationalize  the  denominator  of 


3  +  5V3 


Solve  each  of  the  following  equations 

18.  a*  +  7  ic'  =  8. 

19.  -J— +        ^  ^ 


Vaj+i     Vi— 1     x  —  \ 

20.    Va5  4-5  +  V^a;4-  5  =  12. 


21.    V7a;- 6— A/I7a5  — 2  +  V3a;- 2  =  0. 


22.  \/3a;  +  10+  \/5a=  Vl9ic  +  5. 

23.  2(.-a)^3(.-6)^^^ 

x  —  b  X  —  a 

X       ,         5a;  — 56  a;        ,  7a;+90 

24.  — ;^  +  7 -r-7 — — ^n-  =  — :-—-  + 


a; +8       (a;-4)(a;  +  8)       a;  +  10       (a;  —  6)(a;  +  10) 
X  2  (a;  +  9)       _      x  7^  —  6 


a;  -  3       (ar  +  5)(a;  -  3)       a;  +  6       (a;  -  2)(a;  +  6) 


THE  BINOMIAL  THEOREM  643 


CHAPTER  XXVIII 

THE  BINOMIAL  THEOREM 

1.  In  this  chapter  we  shall  consider  the  laws  governing  the  ex- 
pansion o(  a  binomial  to  any  real  power,  proving  them  for  any 
positive  integral  exponent  and  applying  them  for  positive  or 
negative,  integral  or  fractional  exponents. 

The  Binomial  Theorem  for  Positive  Integral  Exponents 

2.  The  student  should  obtain  the  following  identities  by  actual 
multiplication  : 

Check,  a  =  b  =  1 

ia  +  by  =  a^+       b 2^=   2 

la  +  by  =  a^  +  2ab+         b^ 2^=   4 

la  +  bf=a^  +  Sa''b+   3rWr+         b^ 2«=   8 

(a-\-by  =  a''  +  4a^b+   CyaV/'-h   4a^>«+         b^    .     .     .     .  2^=16 

la-\-bf=a^  +  [)a*b+h)a^b^+Wa'b^+   5ab'+      b^       .  2^  =  32 

(a  +  by  =  a^  +Qa^b+  Ida^b'  +  20a'6«  +  ISr/^^*  +6ab'  +  b"  2'  =  64 

By  inspection  of  the  identities  above,  we  shall  discover  certain 
laws  of  coefficients  and  exponents  which  will  be  found  to  hold  true 
for  the  expansion  of  any  binomial  raised  to  a  power  the  exponent 
of  which  is  any  positive  integer ;  and  they  may  be  shown  to  hold 
without  exception  for  all  powers,  whether  the  exponents  be  positive 
or  negative,  integral  or  fractional,  real  or  imaginary. 

3.  Observe  that,  in  the  expansion  of  a  binomial  to  any  power 
(obtained  for  the  present  by  multiplication)  : 

(i.)  The  number  of  terms  is  greater  by  unity  than  the  index  of  the 
power  of  the  binomial. 

E.  g.  In  the  expansion  of  (a  +  hy  there  are  four  terms  ;  of  («  +  by,  six 
terms  ;  of  {a  +  by,  ten  terms. 


644 


FIRST  COURSE   IN  ALGEBRA 


(li.)  The  expomnts  of  the  first  and  last  terms,  and  also  the  coeffi- 
cients of  the  second  term  and  the  term  next  to  the  last,  are  equal  to 
the  index  of  the  given  binomial. 

E.  g.  In  the  expansion  of  (a  +  h)\  the  number  4  appears  as  exponent 
in  a*  and  6*;  and  as  coefficient  in  4a*6  and  4a6*. 

(iii.)  The  exponent  of  the  first  term,  a,  de^creases  by  unity  in  succes- 
sive terms,  and  the  exponent  of  the  second  term,  b,  increases  by  unity  in 
successive  terms,  appearing  to  the  first  power  in  the  second  term. 

E.  g.  In  the  expansion  of  («  +  6)',  a  has  the  following  exponents  in  the 
successive  terms :  6,  5,  4,  3,  2,  jnul  1 ;  6,  beginning  with  the  second  term, 
has  the  exponents  1,  2,  3,  4,  5,  and  6. 

The  distribution  of  the  exponents  among  the  literal  factors  of  the 
successive  terms  of  the  expansion  of  («  +  bf  is  indicated  in  the 
following  table : 


Terms            1        2 

3 

4 

5 

6 

7 

8 

9 

10 
9 

(a 
Powers  of  ] 
\b 

9 

8 

1 

7 
2 

6 
3 

5 
4 

4 
5 

3 
6 

2 

7 

1 
8 

Sum  of 

Exponents 

9 

9 

9 

9 

9 

9 

9 

9 

9 

9 

The  exponents  of  the  literal  factors  in  the  different  terms  wiU 
accordingly  appear  as  follows  : 

9        81        72        63        54        45        36        27        18       9 

Inserting  the  letters,  we  have  the  following  powers  of  a  and  b  : 
a\  a^b\  a^b\  a%\  a'b',  a^b\  a''b\  a^b\  a^b\  b\ 

For  the  expansion  of  {a  +  b)  to  any  power,  such  as  the  nih,  we 
have 

a",  a"-^&,  a"-^^^  «"-'^^  a"-*^^  a''-^^ ,  a%''-\  ab''-\  b\ 

(iv.)  The  sum  of  the  exponents  oj  a  and  b  in  any  term  of  a  given 
expansion  is  equal  to  the  index  of  the  power  of  the  binomial. 

E.  g.  As  indicated  in  the  lowest  row  of  the  table  for  the  exponents  in  the 
expansion  of  the  ninth  power  of  (a  +  6)  given  above,  the  sum  of  the  ex- 
ponents of  the  literal  factors  in  each  term  is  9. 


THE   BINOMIAL  THEORExM  645 

(v.)  The  coefficient  of  the  first  term  of  every  expansion  is  1,  and 
that  of  the  second  term  is  equal  to  the  index  of  the  power  to  which  the 
binomial  is  raised, 

E.  g.  Ill  the  expansion  of  (a  +  by,  the  coetficient  of  the  first  term  is  1, 
and  that  of  the  second  term  is  4. 

We  have  the  following  multiplication  and  division  rule  for  calcu- 
lating the  coefficients  successively : 

(vi.)  In  the  expansion  of  the  binomial  {a  +  b)  to  any  power,  we 
may  find  the  coefficient  of  any  specified  term  from  the  coefficient  and 
exponents  of  the  preceding  term  by  dividing  the  product  of  the  coefficient 
and  exponent  of  the  first  letter,  a,  in  this  preceding  term,  by  a  num- 
ber which  is  greater  by  unity  than  the  exponent  of  the  second  letter,  b. 

E.  g.  In  the  expansion  of  (a  +  6)®,  we  may  obtain  the  coefficient  15  of 
the  third  term,  15  rt*^^,  from  the  coefficient  and  exponents  of  the  second 
term,  Qa%,  by  multiplying  the  coefficient  6  by  the  exponent  5  and  dividing 
the  result  by  a  number  greater  by  unity  than  the  exponent  of  6,  —  that  is, 
by  2. 

Similarly,  we  may  obtain  the  coefficient  20  of  the  fourth  term,  20  a%^, 
from  the  coefficient  and  exponents  of  the  third  term,  15  a^ft^,  by  performing 
the  operations  15  x  4  -f-  3  =  20. 

To  find  the  coefficient  15  of  the  fifth  term,  15  a%\  we  use  the  coefficient 
and  exponents  of  the  fourth  term  as  follows  : 

20  X  3  -f  4  =  15. 

The  coefficient  of  the  sixth  term  is  15x2-f5=:6. 
To  obtain  the  coefficient  of  the  seventh  term,  we  write 

6  x  l-f6=  1. 

If  we  attempt  to  calculate  the  coefficient  of  another  term  after  the  seventh, 
we  shall  have  1  x  0  ^  6  =  0,  and  since  the  coefficient  is  zero,  no  such  term 
exists. 

In  calculations  such  as  these,  it  is  convenient  to  perform  the 
divisions  before  the  multiplications. 

E.  g.  In  calculating  the  coefficients  in  the  expansion  of  (x  +  yy^  we 
may  begin  by  writing  (x  -{- yy°  =  x^^  +  20x^^y  + 

To  calculate  the  coefficient  of  the  third  term,  we  may  proceed  as  follows : 

20  X  19  -^  2  =  20  -^  2  X  19  =  10  X  ]  9  =  190. 
Hence,  the  third  term  is  IQOx^^y^ 


646  FIRST  COURSE  IN   ALGEBRA 

To  obtain  the  coefficient  of  the  fourth  term,  we  may  write 

190  X  18  -^  3  =  190  X  6  =  1140. 
Hence,  the  first  four  terms  of  the  required  expansion  of  (x  +  ij)^,  are 

(x  +  2/)20  =  x20  +  20  x^^y  +  190  x^^y^  +  1 1 40  x^''y^  + 

The  student  should  complete  the  expansion. 

(vii.)  The  coefficients  are  repeated,  in  inverse  order  after  passing 
the  middle  term  or  terms  oj  any  expansion^  so  that  after  the  coeffi- 
cients of  the  terms  in  the  first  half  of  any  particular  expansion  are 
calculated,  the  coefficients  of  the  terms  in  the  second  half  may  im- 
mediately he  icritten. 

(viii.)  If  —h  is  substituted  for  b  throughout  the  expansion  of 
{a  +  by,  the  signs  of  all  terms  containing  odd  ptot^^rs  of  b  will  be 
changed  from  +  to  — ,  and  those  containing  even  powers  of  b  will 
remain  unchanged. 

E.  g.    (a  -  hy  =  a*  -  5  a*6  +  10  a%'^  -  10  aHfi  +  5  «6*  -  h^. 

4.  From  the  observations  above  we  obtain  the  following  rule  for 
expanding  any  power  of  a  given  binomial  (restricting  the  exponents, 
for  the  present,  to  positive  integral  numbers)  : 

In  the  expansion  of  (a  ■\-  by,  the  index,  n,  of  the  given  binomial, 
the  exponent,  n,  of  the  first  term  of  the  expansio7i  (in  which  the  first 
letter  a  appears  alone),  and  the  coefficient,  n,  qf  the  second  term,  are 
equal. 

The  exponent  of  the  first  letter,  a,  decreases  by  unity  in  succeeding 
terms,  while  the  exponent  of  the  second  letter,  b,  increases  by  unity  in 
succeeding  terms,  beginning  with  unity  in  the  second  term. 

To  find  the  coefficient  of  any  term  from  the  coefficient  and  exponents 
of  the  preceding  term,  divide  the  product  of  the  coefficient  and  the  ex- 
ponent of  the  first  letter,  a,  in  this  preceding  term,  by  a  number  greater 
by  unity  than  the  exponent  of  the  second  letter,  b,  in  this  term. 

The  calculation  of  successive  terms  will  stop  when,  in  the  computa- 
tion, a  term  appears  having  a  coefficient  equal  to  zero. 

If  the  given  binmnial  is  a  sum,  represented  by  (a  +  b),  the  signs 
of  all  of  the  terms  in  the  expansion  will  be  positive.  If  the  given 
binomial  is  a  difference,  represented  by  (a  —  b),  the  signs  of  all  of  the 
terms  in  the  expansion  will  be  alternately  positive  and  negative,  those 
terms  being  negative  in  which  odd  powers  of  b  are  found. 


THE   BINOMIAL   THEOREM 


64T 


5.  If  the  given  binomial  is  symmetric,  the  expansion  of  any  power 
of  the  binomial  will  be  S3rmmetric.  Hence,  the  coefficient  of  the  first 
term,  the  coefficient  of  the  second  term,  the  coefficient  of  the  third 
term,  etc.,  will  be  respectively  equal  to  the  coefficient  of  the 
last  term,  the  coefficient  of  the  term  next  to  the  last,  the  coefficient 
of  the  term  which  is  second  from  the  last,  etc. 

Ex.  1.    Expand  (2  a:  +  3?/)^ 
We  have, 
(2x  +  3?/)5  =  (2a:)5  +  5  (2x)^  {3  i/)  +  10(2x)3  (3^)2+10(2x)2  (3.y)3+5  •  2x{3y)*+(3y)^ 
=  32a:S  +240x*y  +720a;V         +1080j;2_y3      -|-810:ry/4     +243,y5. 

Check.   X  =  ?/  =  1. 
3125  =  3125. 
Ex.  2.    Expand  (m^  -  4  n^)». 

We  have,  (m*  -  4  n^y  =  (m^)^  -  3  (m^y  (4  n^)  +  3  (m2)(4  n^y^  -  (4  n^y 
=  m«      -Um^n'^  -\- 48  m^n^  -   64  w^. 

Check,   m  =  3,  ?i  =  1. 
125=  125. 

Proof  of  the  Binomial  Theorem  for  Positive 
Integral  Exponents 

6.  We  have  found  by  actual  multiplication  that  for  positive 
integral  values  of  w,  equal  to  or  less  than  6, 


(a  +  hy  =  a""  +  na^'-^b  + 


+ 


71  (n  —  1) 


a^'-'h' 


n(n—  l)(y^  —  2) 
2-3 


W  + 


(1) 


It  will  now  be  shown  that  the  formula  above  applies  for  all  posi- 
tive integral  values  of  n. 

Multiplying  both  members  by  a  +  ^,  we  have 


+ 


+ 


n{n  —  1) 


+ 
+ 


a"-2^>«+. 


n(n-\){n—2) 
2-3 

2 

2 


(2) 


(3) 


648  FIRST   COURSE   IN  ALGEBRA 

It  may  be  seen  that,  wherever  n  appears  in  (1),  w  +  1  appears  in 
(3).  It  follows  that,  if  the  theorem  be  true  for  any  particular 
integral  number  represented  by  n,  it  will  be  true  also  for  the  next 
higher  value,  that  is  for  the  number  represented  by  n  +  1. 

Hence  the  laws  which  apply  to  the  coefficients  and  exponents  in 
the  expansion  of  {a  +  by  apply  also  to  the  coefficients  and  ex- 
ponents in  the  expansion  of  {a  +  ^)"''"^. 

We  have  found  by  actual  multiplication  that  they  apply  for  the 
second  power ;  hence,  by  the  reasoning  above,  they  must  apply  for 
the  third  power. 

Since  they  hold  for  the  third  power,  they  must  hold  for  the 
fourth  power,  and  so  on  indefinitely. 

Hence,  the  general  formula  (1)  applies  for  all  positive  integral 
values  of  the  exponent. 

The  method  of  reasoning  employed  in  this  proof  is  known  as 
matbeuiatical  induction. 

Exercise  XXVIII.     1 
Write  the  expansion  of  each  of  the  following  powers  : 

1.  im  +  n)\  8.  Qi  -  yf.  15.  (2  +  e)\ 

2.  it  +  ny.  9.  ir  -  sy.  16.   (3  +/)^ 

3.  {d  +  gy.  10.  {p  -  qy\  17.  (2  -  gy. 

4.  (a  +  cy.  11.  (a  +  1)*.  18.  {\  +  v  y. 

5.  {k  +  wy\  12.  {b  -  1)^  19.  (2  a;  -  yy. 

6.  [c  -/y.  13.  (c  -  by.  20.  (2  h  +  wy, 

7.  (6  -  x)\  14.  {d  -  9)2.  21.  (m  +  2  ny\ 

Write  the  first  five  terms  of : 

22.  (a  +  6)^.  24.  (Jc  +  z)"^,  26.  {m  +  xy\ 

23.  {h  +  c  y\  25.  {d  +  r  y\  27.  {v  -  yy\ 
Write  the  first  six  terms  of : 

28.  (^  -  d)^\  29.   (s  -  xy.  30.  (a  -  w^. 

Write  the  first  and  last  three  terms  of : 
31.   (c  +  dy.  32.   (x  +  yy\  33.   (p  +  qy\ 

Write  the  expansion  of  each  of  the  following  : 

34.  (5 a  +  4 by.  36.   (3 w?  -    5 ny.         38.   (2b  -  ^ x)', 

35.  (6  a  -  7  by.  37.  (9  c   -  11  d)\        39.  (4  a;  -  y^. 


THE   BINOMIAL  THEOREM  649 

The  Binomial  Theorem  for  Negative  or 
Fractional  Exponents 

7.  It  may  be  proved,  by  means  of  certain  principles,  that  the 
binomial  theorem  applies  even  when  the  exponent  is  a  negative  or  a 
fractional  number. 

Without  proving  the  theorem,  we  shall  assume  that  for  all  positive 
or  negative,  integral  or  fractional  values  of  the  exponent,  the  formula 
is  true  in  the  following  form  for  such  values  of  the  letters  as  make 
the  first  term  of  the  given  binomial  greater  than  the  second  term  : 

(a  +  by  E  a"  +  7^^"-'^>  +  ^^^  ~  ^K^-%^ 

+  "(''-/)("-'^>^»-w+ ., 

It  may  be  seen  that,  whenever  n  is  negative  or  fractional,  the 
numerator  of  the  coefficient  of  the  rth  or  general  term, 

n(n  —  l)(n  -  2) (n  —  r  +  2) 

1  •  2  •  3 (t"  1)    ' 

can  never  become  zero,  and  hence  the  number  of  terms  for  a  parti- 
cular expansion  is  unlimited. 

Ex.  1.   Write  the  expansion  of  (2  a^  -  3  b~^y. 

We  have, 

(2a^  -  36-^)»  =  (2a^)8  -  3(2a^)2(3r^)  +  3(2a^)(3ri)2_  (36  "1)8; 

=    8a      -36ah~^  +54a*fe-i  _27  6~^- 

Check.     Let    a  =  8,     b  =  4. 

Ex.  2.   Write  the  first  four  terms  in  the  expansion  of  (1  —  2  x)~^. 
We  have, 

2  ^  ■  o 


=  I  +4a:+  12a:2  +  32a;8  + 


Since  (1-2  a;)-2  =  - — L-—  = -i — ^ , 

^  (l-2ic)2      l-4a;  +  4x2' 

we  can  check  the  result  by  performing  the  division 

j__2__^=  1+4.=  + 120.^  + 32.3  + 


650  FIRST  COURSE   IN  ALGEBRA 

Ex.  3.    Write  the  first  four  terms  in  the  expansion  of  (8  —  x)~^. 
We  have, 


I       X        x^         7x» 
-  2  ^  48  ^  576  ^  41472  ^ 


Since  (8  —  ar)   ^  =  — L L ,  jt  is  possible  to  check  the  result  by  divid- 
ing ^^64-  16x4- x»  by  8  -  a:. 

Exercise  XXVIII.     2 
Write  the  expaDsion  of  each  of  the  following  : 

1.  (a-^  +  ^»)^  6.  (772-2  _^  2  w)*. 

2.  (3  cr-^  -  0^)2.  7.  (r'-2  +  3  ^/e^)^ 

3.  (5a^ij-^c-Ax~^i/-y.  8.  (a^"^-7/-^)^ 

4.  (a-^  +  6*)«.  9.  (</*  -  k^y. 

Expand  to  four  terms  each  of  the  following  : 

11.  (1  +x)-\  19.  (1  +  x)K 

12.  (1  —  a)-\  20.  (1  4-  7/)i 

13.  (1  -  b)-^  21.  (1  +  z)^. 

14.  (1  —  c)-\  22.  (1  —  m)~K 

15.  (l-d)-^.  23.  (4-3ic)"l 

16.  (l-^)-'.  24.  (l-3r)~l 

17.  (l-2n)-\  25.  (l  +  4^)"t 

18.  (a-^/2)-6,  26.  (8-^3a;)l 


THE   BINOMIAL  THEOREM  651 

Selection  of  a  Particular  Term  in  the  Expansion  of 

{a  +  6)" 

8.  Consider  the  following  terms  in  the  expansion  of  {a  +  ^)",  n 
being  positive  and  integral. 

{a  +  by  =  «»  +  noT-^b  4-  ^^^-^  «""'/>' 

^  .(.-!)(.  -2)  ^..3,3^ 

It  should  be  observed  that  the  denominator  of  the  coefficient  of 
any  particular  term  consists  of  the  product  of  the  primary  numbers 
1,  2,  3,  etc.,  up  to  a  number  which  is  less  by  unity  than  the  number 
of  the  term. 

E.  g.    In  the  fourth  term  the  denominator  is  1  •  2  •  3. 
Accordingly,  the  tienominator  of  the  coefficient  of  the  rth  term  from  the 
beginning  must  consist  of  the  product  1  •  2  •  3  •  4  •  5 (r  —  1). 

9.  The  product  of  the  successive  primary  numbers  1,  2,  3,  4,  5, 

,   up   to  and  including   any  specified  number,  /,  is  called 

factorial  /,  and  may  be  indicated  by  |_  or  !,  as  shown  in  the  fol- 
lowing illustrations  : 

E.g.   If/ be  5,  [5  =l-2-3-4-5  =  120. 

4  !  =  1  •  2  •  3  •  4      =24. 

a\  =  1-2-3-4 (a-2)(a-l)a. 

(r  -  1)  !  =  1  •  2  •  3  •  4 0  -  2)(r  -  1). 

10.  The  denominator  of  the  coefficient  of  any  specified  term  in 
the  expansion  of  a  binomial  to  any  power  can  be  expressed  by  the 
fctctorial  notation. 

E.  g.  In  particular,  the  first  five  terms  in  the  expansion  of  (a  +  hy^  may 
be  written  as  follows  : 

^a+hy^=a^o^\Oa%+^-^^aW^^^a^^^^  • 

Or  («+6)io=aio  +  iOa«6+^a«6Hi^a'68+^^^^^^ 

11.  By  examining  any  particular  term  in  the  expansion  of 
{a  4-  by  (see  §  8)  it  may  be  observed  that: 


652  FIRST  COURSE   IN  ALGEBRA 

(i.)  The  exponent  of  the  second  letter^  6,  is  equal  to  the  last  factor 
in  the  denominator  of  the  coefficient. 

E.  g.  In  the  fourth  term  the  exponent  of  the  second  letter  b  is  3,  which 
is  the  hist  ftictor  in  the  denominator  of  the  coefficient. 

In  the  rth  term,  the  last  factor  in  the  denominator  is  r  —  1,  and  the 
exponent  of  the  second  letter  6  is  r  —  1. 

(ii.)  The  exponent  of  the  first  letter,  a,  is  eqiml  to  the  difference 
between  the  ind^x  of  the  power  to  which  the  given  binomial  is  raised 
and  the  exponent  of  the  second  letter,  6,  in  the  same  term. 

E.  g.  In  the  expansion  of  (a  +  6)"  the  exponent  of  a  in  the  rth  term  from 
the  b^inning  is  7i  —  (r  —  1)  =  w  —  r  +  1. 

(iii.)  The  first  factor  in  the  numerator  of  any  coefficient  is  n  which 
is  the  index  of  the  power  to  which  the  given  binomial  is  raised.  The 
successive  factors  w  —  1,  w  —  2,  etc.,  decrease  successively  by  unity, 
and  the  number  of  factors  in  the  numerator'  of  any  coefficient  is  equal 
to  the  number  of  factors  in  the  denominator. 

E.  g.  In  the  expansion  of  (a  +  hy^  the  fii-st  factor  in  the  numerator  of 
each  numerical  coefficient  is  10,  and  in  each  coefficient  the  number  of 
factors  in  the  numerator  is  equal  to  the  number  of  factors  in  the  denomina- 
tor. It  should  be  observed  that  the  last  factor  in  each  numerator  is  greater 
by  unity  than  the  exponent  of  the  first  letter,  a,  in  the  same  term. 

In  particular,  in  the  fourth  term  the  last  factor  in  the  numerator  is  8 
which  is  greater  than  7  by  unity ;  in  the  fifth  term  the  last  factor  in 
the  numerator  is  7  which  is  greater  than  6  by  unity. 

Ex.  1.  Write  the  8th  term  in  the  expansion  of  (a  +  h)^^. 

Since  all  of  the  terms  in  the  expansion  of  a  binomial  sum  are  posi- 
tive, the  sign  of  the  8th  term  must  be  positive. 

Since  the  exponent  of  the  power  to  which  the  second  letter  h  is  raised  is 
less  by  unity  than  the  number  of  the  term,  the  exponent  of  h  in  the 
8th  term  must  be  7.  Accordingly,  the  coefficient  of  a,  which  is  found 
by  subtracting  the  exponent  of  h  from  the  index  of  the  power  to  which  the 
given  binomial  is  raised,  must  be  10  —  7  =  3. 

To  obtain  the  numerical  coefficient  we  may  write  a  fraction  the  numera- 
tor of  which  consists  of  the  product  of  the  seven  numbers  10  •  9  •  8  •  7  •  6  •  5  •  4, 
and  the  denominator  of  which  is  7 !,  that  is,  the  product  l-2-3'4-5-6-7. 

Hence  the  required  8th  term  is 

10  •  9  ■  8  ■  7  •  6  •  5  •  4 
+    1-2-3-4-5-6-7"" 


THE   BINOMIAL   THEOEEM  653 

12.  It  may  be  seen  that  the  rth  term  or  general  term  in  the 
expansion  of  (ci  +  hy  may  be  written  as  follows  : 

n{n  -  \){n  -  2){n  -  3) {n  -  r  ■{- 2)  ,_, 

1    •     2     •     3     •     4     (^  _  1)     "        ^     • 

13.  It  should  be  observed  that  in  the  expansion  of  {a  —  hf  the 
sign  of  every  term  is  negative  in  which  the  exponent  of  the  power 
to  which  —  6  is  raised  is  odd,  and  the  sign  of  every  term  is  positive 
in  which  the  exponent  of  the  power  to  which  —  6  is  raised  is  even. 
Accordingly,  since  in  the  rth  term  the  exponent  of  the  power  to 
which  —  b  is  raised  is  r  —  1,  it  follows  that  the  sign  of  the  term  may 
be  determined  by  the  factor  (—  1)'"^ 

Hence,  we  have  the  following  expression  as  the  rth  term  or  general 
term  in  the  expansion  of  {a  —  by  : 

/     ,.,._,  n{n-\){n-2){n-^) (y^  _  ^  +  2)  , 

C-1)       1    .  2      •       3      •      4    ir-l)'' 

E,  g.  In  the  expansion  of  (x  —  yY^  the  fifteenth  term  containing  (—  yy^ 
is  positive ;  the  sixteenth  term  containing  (—  yY^  is  negative.  In  the  ex- 
pansion of  (m  —  n)"  the  nineteenth  term  containing  (—  ny^  is  positive;  the 
twenty-sixth  term  containing  (—  nY^  is  negative. 

Ex.  2.    Write  the  sixth  term  in  the  expansion  of  (h  —  c)i^. 

We  have  (-  l)«-i  \^  '  ^f  '  V  '  \^  '  ^l  ^''''  =  "  11628  b^'c^ 
1'2'3'4*5 


Exercise  XXVIII.    3 

Write  the  indicated  terms  in  the  expansions  of  the  following 
powers  of  binomials  : 

1.  The  5th  term  of  (a  +  by\  8.  The  11th  term  of  {p  -  qY'. 

2.  The  8th  term  of  {b  +  cy\  9.  The  10th  term  of  {a  -  yy\ 


3.  The  7th  term  of  (c  +  dy\  10.  The    4th  term  of  {g  —  h) 

4.  The  9th  term  of  {x  +  yy\  11.  The  13th  term  of  (c  -  wY^. 

5.  The  6th  term  of  (m  +  nY\  12.  The  20th  term  of  (z  +  wY^ 

6.  The  10th  term  of  (r  +  sY^-  13.  The  16th  term  of  (a  +  by^ 

7.  The  15th  term  of  (a  -  by.  14.  The  Uth  term  of  (b  -  cY^ 


654  FIRST  COURSE   IN   ALGEBRA 

Expand  each  of  the  following  powers  of  binomials  : 

15.  (a^  +  hy.  19.  (z'  -  ly. 

16.  (c^-d)\  20.  (2a+by. 

17.  (£c'-2)*.  21.   (2a»+  8/^V. 

18.  (f  —  1)^  22.  (37W^  +  5  ?iy. 

14.  The  formulas  of  §§  12,  13,  for  the  general  or  rth  term  in 
the  expansion  of  (a  +  by  and  (a  —  by  may  be  used  also  when  the 
exponent  n  is  negative  or  fractional. 

Ex.  1.  Find  the  prime  factors  of  the  coefficient  oia^lr'^^  in  the  expan- 
sion of   («-2^)    • 

It  may  be  seen  that  a^^  and  6"^^  appear  in  the  term  of  the  expansion  in 
which  a^  and  —  (ot)      ^^^  found. 

Since  the  exponent  of  the  power  to  which    (  —  ^)    is  raised  is  less  by 

unity  than  the  number  of  the  term,  it  appeal's  that  we  are  required  to 
write  the  12th  term. 

Since  the  given  binomial  is  a  difference  and  in  the  12th  term  the  factor 

( —  —  j    is  raised  to  the  11th  power,  it  may  be  seeu  that  the  sign  of  the 

term  is  minus. 

Hence,  we  may  write 

31  .  30  .  29  .  28  •  27  •  26  •  25  .  24  .  23  .  22  .  21 


25  >  24  »  23  '  22  •  21    2o/J_V^- 
7-    8-    9-  10-  11*    Ylb)     - 


1  .    2  .    3 •    4-    5 •    6 

-  (^)iA  .  3»  •  5  .  7  •  13  •  23  •  29  •  31  a^b-^^. 
Ex.  2.    Write  the  term  containing  x^  in  the  expansion  of 


("-;<)" 


Referring  to  the  general  formula  for  the  rth  term  in  the  expansion  of 
(a  —  6)",  §  13,  it  may  be  seen  that  x^  appeiiring  in  the  first  term,  '2x^,  of 
the  given  binomial,  is  to  be  raised  to  the  power  represented  by  (m  —  r  +  1) 
in  the  formula  for  a  general  term,  and  that  x~^  appearing  in  the  second 
term,  [ j-  ],  of  the  given  binomial,  is  to  be  raised  to  the  power  repre- 
sented b  y  r  —  1- 

When  reduced  to  simplest  form,  the  rth  term  must  contain  x^. 


THE  BINOMIAL  THEOREM  655 

Hence,  observing  that  n  =  17,  we  may  find  r  as  follows : 

Hence  a;    2      =  x^^. 

Since  the  bases  are  equal  we  can  form  a  conditional  equation  of  which 
the  members  are  the  exponents. 

That  is,  lOQ-lr  ^  ^^^ 

Hence,  r  =  7. 

Accordingly,  we  are  required  to  find  the  7th  term,  which  may  be  written 
as  follows: 


H-^f^^^f^^f^(-)"gr-— — ^ 


a:i 

=  16384-  729-  1547  a;^ 
=  18477268992x80. 


Exercise  XXVIII.     4 

Write  the  specified  terms  in  the  expansions  of  the  following 
powers  of  binomials  : 

1.  The  4th  term  of  («-'  +  h^Y\ 

2.  The  6th  term  of  (6"^  +  2  t^y. 

3.  The  8th  term  of  (w"^  +  4  n-y\ 

4.  The  13th  term  of  (a;  —  xhj)'^. 

5.  The  term  of  {x^  +  2  cc"^)^^  which  contains  a;". 

6.  The  middle  term  of  («"«  +  "lahyK 

7.  The  term  of  {^  —  y^Y'  which  contains  x^y^. 

8.  The  term  of  (3  a;  +  2  a?"^)^"  which  is  free  from  a;. 

9.  The  5th  term  of  (1  -  2  a;)^. 

10.  The  7th  term  of  (1  -  x^^. 

11.  The  5th  term  of  («-*  -  ^'-')-'- 

12.  The6thtermof(a^-^>-V)-«. 


656  FIRST  COURSE   IN  ALGEBRA 

15.  The  coefficients  appearing  in  the  expansion  of  (a-\-by,  called 
biiiouiial  coefficients,  may  be  denoted  by  the  following  abbrevi- 
ations : 

_n__fn\  n  {n  —  \)  _  fn\         n  (n  —  l)(n  —  2)       fn\ 

'^^l-Vl/  1-2       -{2}  l^J^S  -\S/ 

n  (n—  l)(n  —  2) (n  —  r  +  2)  _  f     7i     \ 

1-2-3 (^—1)  ~\r-l)' 

Accordingly,  we  may  write 

It  may  be  observed  that  in  the  notation  above  the  upper  number 
in  each  symbol  is  the  exponent  of  the  power  to  which  the  given 
binomial  is  raised,  and  the  lower  number  indicates  the  number  ot 
factors  appearing  in  both  numerator  and  denominator  of  the 
numerical  coefficient. 

E.  g.  In  the  expansion  of  (a  +  by^  the  coeflBcient  of  the  7th  term  is 
/10\       10-9-8-7-6.5 

In  the  expansion  of  (a  4-  6)*  the  coefficient  of  the  fifth  term  is  I  j  = 
4. 321       .  ^^^ 


1.2-3.4 


=  1. 


16.   The  binomial  theorem  may  be  used  to  obtain  any  required 
root  of  a  number,  as  illustrated  in  the  following  example  : 

Ex.  1.   Find  the  square  root  of  11  correct  to  four  places  of  decimals. 
We  may  write  //H  =  ^/3^ -\- 2  =  (3^  +  2)i. 

Hence,  to  find  yTT  we  may  expand  (3^  +  2)i  as  follows: 

(32+2)i=(32)^+^.32)-*-2-i(3T^-2^-h^(3T^  '  2»-ifg<32)-^  '2*+ , 

=  3  +  i  —  -g^T  +  J^  ~  TTf57  + » 

=  3  + (.33333+) -(.01851+) +  (.00205+) -(.00020+)  + 

=  3.3166  +  ,  correct  to  four  places  of  decimals. 

Exercise  XXVIII.    5 
Find  to  four  places  of  decimals  : 

1.  V26.  4.   v^.  7.   V^n.  10.   v^65. 

2.  \/50.  5.   v^.  8.   v^.  11.   v^730. 


3.   a/102.  6.   Vi24.  9.   a/626.  12.   V2188. 


EEVIEW 


657 


Mental  Exercise  XXVI 1 1.     6.    Keview. 
Solve  each  of  the  following  equations  : 

'     "       ^      0. 


1. 

1     t)' 

a~  ^' 

4.  \^^x'-b^  = 

5.  \^x  -\-  b  =  a. 

2. 
3. 

mot? 
n 

6.  f  -  a'b. 
b 

=  mn. 

8. 

Show  that  a;*  +  //  = 

:0'\ix-\-b  =  0. 

;in 

iplify  the  following : 

o 

y-'^ 

2-x 

10. 


V  y  —  3  '  Vx  —  2 

11.  (a/5  +  a/=^)(a/5  -  ^=^3). 

12.  (V^y  +  a/-'3)(a/^^  -  a/=^). 

13.  {a^  +  i)l 


14.  (b^  -  ^y. 

15.  (2xi  +  ^y. 

\x      y       zj\xy  -\-  yz-^  zxj 


16.  (3\/- 3  -  3a/3)''. 


b's/a 


18.  ^' 


20.  -  (-  a;V^^)^ 
Find  the  value  of : 

21.  ix'^  +  fy^ 

22    A. 

25.    Show  that  -  3\ 


10 


24.   (-  V2)^ -( V-  2)2_. 
3  =  3V^3  while  -  2a/-  2  7^  2^2. 


26.  Simplify    -( -  )    ,    and    express   the    result   with   positive 
exponents. 

27.  Show  that  a,  b,  c,  and  d  are  proportional  if  7  :  -  =  1. 

28.  Express  2\^\/h  in  simplest  form. 


42 


658  FIRST  COURSE   IN  ALGEBRA 

Simplify 
29.    (?i^^')(fi  -  l)Vn.  30.    (^l  +  c-'^('^^-cA' 

31.  Find  the  geometric  mean  oi  a  and  I /a. 

32.  Find  the  mean  proportional  between  a  and  1/a. 

Classify  each  of  the  following  progressions  as  being  arithmetic 
harmonic,  or  geometric  : 

33.  2,  4,  6, 34.    2,  4,  8, 35.    i,  i,  ^ 

Find  the  values  of : 

3'  8' 

36.    2!  3!.  37.   ^.  38.   |j. 

Show  that  the  following  identities  are  true  : 

39.  4  !  =  3  !  2  !  2  ! .  41.   ^  +  2  •  3  !  =  3  •  4  ! . 

95         1  1 

40.  2  •  3  !  -  3  '  2  !  =  3  ! .  42.    f^  =  -^  +  ^ . 

5  1       4  !       3  ! 

Exercise  XXVIII.     7.    Review 

Simplify  each  of  the  following  expressions  : 

1.  6a/yc4-(c  +  a)*+  (^  +  c)« +  («  +  />)'-(«  + ^  +  c^. 

2.  («+«/  +  zy  ^{x^.y-zf-(^^rz-  xf-iz  ^-x-  y)\ 

1 


3. 


^—^-' 


.+   1 


iC+  1 


a  —  h  h  —  c  c  —  a 


6.  Showthat(-^+-^  +  -^Vf-^+A  +  -^^  =  3, 

\x—a      x—o      x—cj      \xr—a      x—b      Xr-Cj 

\i  X  i^  a,  X  ^  b^  X  ^  c. 

7.  Show  that  [{a  -  by  +  (b  -  cy  +  (c  -  ayf 

=  2  [(a  -  by  +  (/>  -  c)*  +  (c  -  a)^]. 


EEVIEW  659 


x--y 


-2 


8.  Simplify     ~—^ ^ ,     using  the  minimum  number  of  nega- 

X         y 

tive  signs. 

9.  Simplify  ^  X  g^'  X  g. 


10.   Simplify 


4"  •  2  X  ^  -  32' 

2'"  X  4 


12.   Show  that     - 


k'*i)k'-ii 


(-«-)"('-^) 


iv  -\b) 


13.   Show  that 


(-:)"0-f) 


14.  For  what  value  of  n  is  a;"'^^?/"'''^  —  t/^"^""  ^  a  homogeneous 
binomial  % 

15.  For    what  value    of  n  is   a;""'"'*^^  +  ^"^2+^  a   homogeneous 
binomial  ? 

16.  Simplify  "li^^:^^- 

17.  Square  the  complex  number  —  \  —  \y—  2. 


18.  Simplify  (\/m  +  ^  +  Vm^—nY  —  (\/mr^i^  —  Vm  —  rif. 

19.  Simplify         LV  '^  V  '^         J 

3  —  2V3  _ 

V30 

20.  Rationalize  the  denominator  of 


a/S  +  V3  -  a/2 
Solve  each  of  the  following  equations  : 

21.  \/«  +  2  +  V'4  a;  —  3  —  a/D  £c  +  1  =  0. 

22.  (a^  +  -)    -4  fa3+  ^  j  ==  60. 


600  FIRST  COURSE  IN  ALGEBRA 

oQ        g       ,  8a;  — 77         _       x  11a;— 70 


a;  +  7       {x—  12)(a;  +7)       a;  -H  10       (ic  —  H){x  +  10) 

X       .         8a;  — 35                  x        .         4a;  — 99 
24'    ?  +  7 :7w TT  =  7 ^  + 


27. 


x—b       (a;  — 6)(a;  — 5)       (a;  —  9)       (a;  —  2)(a;  —  9) 

„.    _x 5  ^      X 2(a;  +  9) 

•a;-l       (a;+4)(a;— 1)       a;  —  3       (a;  +  r>)(a;  -  3)  * 

2^        a?       ,  g—  12         __  __^ 2(a;  +  5) 

*  a;  -  2       (a;  +  3)(a;  -  2)  ""  a;  -  5       (a;  -  l)(a;  -  5)* 

X  2a;— 15        _      x 3  a;—  16 

a:  -  3       (a;  —  iS){x  —  3)  ~  a;  -  4       (a;  —  5)(a;  -  4)  ' 

X  3(2 a; -33)     _  _x 270 

a;  -  9  "^  (a;  -  4)(a;  -  9)  ~  a;  -  15       (a;  +  3)(a;  -  15)  * 
Solve  the  following  systems  of  equations  : 

29.  18  a;  +  12y=  1  xy, 

12a;  +  18^  =  8a-j^.  31.  7^  +  xy  +  y  =21, 

y\-m       n  x   +  xy  +  y' =    <d. 

30.  X  =  '^———  +  X . 

^  3  32.  a;'  _  ic^  _  7/   =  22, 

„  =  ^-±^  +  ^.  ^^  -a;^  +  /  =  -2. 

^  2  3 

33.  Expand  (3  a;"*  —  'ly^f  by  the  binomial  theorem. 

34.  Expand  ( 1  —  i  x^y  by  the  binomial  theorem. 

35.  Find  the  middle  term  in  the  expansion  of  f  a; J  • 

36.  Find  the  middle  term  in  the  expansion  of  (  ^^'^  +  ^  )    * 

(<i      xY 

37.  Find  the  middle  term  in  the  expansion  of  I  -  H —  J  • 

38.  In  the  expansion  of  f  a;  +  -  j    write  the  term  which  is  free 
from  X, 

39.  Find  the  term  free  from  x  in  the  expansion  of  (  a; j    • 

40.  Find  the  coefficient  of  a^%^  in  {a  +  hy\ 


INDEX. 


Numbers  Refer  to  Pages. 


Abscissa,  140. 

Antecedent,  594, 

Associative  Law  for  addition,  24. 

for  division,  62. 

for  multiplication,  52. 
Axes  of  Reference,  138. 
Axiom,  9. 
Axis  of  imaginary  numbers,  457. 

of  real  numbers,  457. 

Base,  54. 
Binomial,  71. 

square  of,  102,  103. 

coefficients,  656. 
Binomial  Theorem,  111,  646. 

extraction  of  roots  by,  655. 
Boyle's  Law,  606. 

Cancellation,  258. 
Charles'  Law,  607. 
Checks,  numerical,  81,  127,  230,  310. 
Clearing  of  fractions,  298. 
Coefficient,  69. 

Commutative  Law  for  addition,  23, 
36. 

for  division,  61. 

for  multiplication,  49,  50. 
Consequent,  594. 
Constant,  1. 
Coordinate,  140. 
Cross  products,  108. 
Cube,  5. 

Cube  root,  72,  366. 
Cube  roots  of  unity,  545. 

Definition,  24. 

Degree,  70,  154,  229,  231,  245. 


Denominator,  118,  252. 

lowest  common,  262. 
Detached  Coefficients,  82,  98,  127. 
Dimension,  70. 
Discriminant,  514. 
Distributive  Law  for  division,  65. 

for  multiplication,  52. 
Division,  57. 

extended  definition  of,  60. 
Divisor,  greatest  common,  228,  230, 

232. 
Double  signs,  22,  136,  202,  368,  369, 
474,  632. 

Eliminant,  551. 
Elimination,  334. 
Equations,  6,  149. 

binomial,  545. 

conditional,  7,  152,  177,  178,  321, 
327,  359,  503,  505. 

consistent,  330. 

degree  of,  154,  298. 

equivalent,  155,  330,  332,  333. 

fractional,  298,  491. 

graphic  solution  of,  329,  472,  554, 
560. 

homogeneous,  562. 

identical,  149. 

"impossible,"  529. 

inconsistent,  330,  331. 

independent,  330. 

in  quadratic  form,  538,  541. 

irrational,  526. 

linear,  168,  225,  326. 

literal,  306. 

member  of,  6,  149,  152. 

quadratic,  225,  469. 


662 


INDEX 


Equations,  radical  (irrational),  526. 

reciprocal,  545. 

simultaneous,  327,  550. 

symmetric,  572. 

system  of,  327,  550. 
Equivalence,  Principles  of,  156,  157, 
158,  161,  162,  163,  164,  165,  224, 
298,  334,  335,  341,  554,  558,  566, 
570. 
Evolution,  370. 
Expansion,  102,  643. 
Exponent,  4,  54,  69,  391,  392,  393, 

396. 
Expressions,  algebraic,  4. 

equivalent,  6. 

fractional,  256. 

irrational,  72,  415. 

mixed,  256. 

(radical),  72,  415. 

rational,  72. 

Factor,  4,  52,  112,  185,  295,  439, 
462. 

common,  228. 

highest  common,  228. 
Factor  Theorem,  132. 
Falling  bodies,  510,  614,  627. 
Finder  term,  194. 
Form,  93. 
Formula,  308,  502." 

association,  89,  391. 

distribution,  88,  89,  391. 
Fractions,  complex,  283. 

continued,  286. 

decimal,  174,  288,  584,  638. 

equivalent,  257. 

improper,  253. 

principles  relating  to,  277,  412. 

proper,  253. 

reciprocal  of,  279. 
Function,  5,  137,  609. 

Graph,  142. 

of  an  equation,  300,  328,  470. 
of  consistent  equations,  330,  331. 
of  equivalent  equations,  332. 
of  fractional  equations,  300. 
of  a  function,  142. 


Graph,  of  inconsistent  equations,  331. 
of  hnear  equations,  328. 
of  quadratic  equations,  470. 
Grapliic  interpretation  of  solutions, 

344,  554,  560. 
Graphic  solutions  of  equations,  329, 
472,  554,  560. 

Homogeneous,  72,  99,  205,  210,  220. 
562.  "• 

Identical,  10,  149. 
Identity,  6,  149. 

converse  of,  10. 
Inclined  plane,  605. 
Indeterminate  Forms,  290,  295. 
Index,  54,  366. 

Indices,  Laws  of,  88,  112,  391. 
Inequality,  42. 

symbols  of,  7. 
Infinitesimal,  293. 
Infinity,  21,  42,  293. 
Integer,  5. 
Involution,  370. 

Known  numbers,  1. 

Law  of  Exponents,  88,  112. 

of  Polynomial  Quotients,  135. 

of  Quality  Signs,  59,  67. 

of  signs  for  addition  and  subtrac- 
tion, 28. 

of  signs  for  multiplication,  48. 

of  signs  for  x  and  -f ,  64. 
Lever,  Horizontal,  323. 
Limit,  291,  637. 
Linear  Expansion  of  Solids,  608. 

Mathematical  Induction,  648. 
Mean,  arithmetic,  622. 

geometric,  632. 

harmonic,  628. 
Modulus,  465. 
Monomial,  71. 
Multinomial,  71. 
Multiple,  common,  245. 

lowest  common,  245. 
Multiplication,  45. 


INDEX 


663 


Multiplication,  extended  definition 
of,  4G. 

Notation,  factorial,  651. 

functional,  5,  348. 

solidus,  2,  58,  252. 
Numbers,  abstract,  34. 

algebraic,  21. 

"artificial,"  458. 

commensurable,  367,  411. 

complex,  458. 

concrete,  35. 

conjugate  complex,  458. 

even,  54. 

finite,  294. 

general,  318. 

imaginary,  370,  446. 

incommensurable,  367,  411. 

irrational,  367,  415. 

literal,  318. 

negative,  18,  251. 

odd,  54. 

positive,  19. 

rational,  367,  415. 

real,  370,  447. 
Numerator,  118,  252. 

Operand,  4. 
Opposites,  14. 
Ordinate,  140. 
Origin,  139. 

Parentheses,  preceded  by  +,  31. 

preceded  by — ,  31. 

preceded  by  x,  62. 

preceded  by  -^,  63. 
Pendulum,  simple,  547. 
Physics,  problems  in,  323,  509,  547, 

605,  614,  627. 
Polynomial,  71. 

arranged,  72. 

complete,  94. 

incomplete,  94. 

integral,  71,  72. 

reduced,  80. 

square  of,  105. 
Power,  54,  69,  366. 

even,  54. 

odd,  54. 


Prime,  185,  228. 

Principle  of  No  Exception,  12,  42,  46, 

67,  283,  294,  466. 
Problem,  174. 

general,  318. 
Product,  4, 

continued,  4,  51. 
Progression,  arithmetic,  620. 

geometric,  630. 

harmonic,  627. 
Proof,  24. 
Proportion,  598. 
Proportional,  fourth,  599. 

mean,  598. 

third,  598. 

Quadrants,  139. 

Quadratic  Equations,  225,  469. 

complete,  470. 

incomplete,  470. 
Quantity,  4. 

commensurable,  595. 

incommensurable,  596. 

negative,  13. 

positive,  13. 

Radical,  415. 

Radicand,  366. 

Ratio,  594. 

Rationalization,  434. 

Reciprocal,  59. 

Reduction,  32. 

Remainder  Theorem,  129,  214,  218. 

Roots,  72,  155,  366,  367. 

extra,  162,  491,  527. 

index  of,  72,  366. 

loss  of,  163. 

principal,  369,  397,  399,  405,  449. 

Science,  problems  in,  183. 
Sequence,  619. 
Series,  623. 

arithmetic,  624. 

geometric,  634. 

harmonic,  628. 
Sign  of  continuation,  21. 

radical,  72,  366. 
Signs  of  quahty,  20,  67. 


664 


INDEX 


Solutions,  156,  170,  223. 

interpretation  of,    321,    503,    504, 
505. 

number  of,  347,  552. 
Square,  4,  55. 

completing  the,  191,  440,  478,  485. 

trinomial,  190. 
Square  root,  72,  366. 
Standard  Form,  93,  158. 
Standard  Quadratic  Equation,  469. 
Steps,  24. 

composition  of,  25. 

resultant,  25. 
Subscripts,  346. 
Substitution,  Principle  of,  9. 
Subtraction,  32. 
Sum,  algebraic,  73. 
Summands,  32. 
Surds,  415. 

binomial,  416. 
conjugate,  430. 

dissimilar,  423. 

entire,  426. 

mixed,  426. 

quadratic,  416. 

similar,  423. 
Symbols  of  deduction,  24. 

of  grouping,  23,  30. 
Synthetic  Division,  130,  215. 
Systems,  determinate,  327. 

equivalent,  333,  334,  335,  341,  554, 
558,  566,  570. 

homogeneous,  562. 

indeterminate,  328. 


Systems,  order  of,  552. 
symmetric,  573. 

Terms,  69. 

constant,  94. 

degree  of,  70. 

like,  71. 

of  a  fraction,  252. 

similar,  71. 

variable,  94. 
Theorem,  24. 
Transposition,  157. 
Trinomial,  71. 
Type,  93. 

Units,  quality,  21. 
Unknowns,  1,  154. 

Value,  absolute,  14,  20,  369,  307. 

approximate,  148,  412,  414,  472, 
481,  486. 

expressed,  306,  551,  566. 

indeterminate,  292. 

numerical,  5. 

of  an  expression,  291. 
Variable.  1,  137. 

dependent,  609. 

independent,  609. 

Zero,  as  an  exponent,  392. 
as  a  factor,  53. 
as  a  power,  393. 
division  by,  58. 
power  of,  55. 


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